【图论 拓扑排序 bfs】P6037 Ryoku 的探索|普及+

本文涉及知识点

C++图论
C++BFS算法

P6037 Ryoku 的探索

题目背景

Ryoku 对自己所处的世界充满了好奇，她希望能够在她「死」之前尽可能能多地探索世界。

这一天，Ryoku 得到了一张这个世界的地图，她十分高兴。然而，Ryoku 并不知道自己所处的位置到底在哪里，她也不知道她会什么时候死去。她想要知道如何才能尽可能多的探索这个世界。

题目描述

Ryoku 所处的世界可以抽象成一个有 $n$ 个点， $n$ 条边的带权无向连通图 $G$。每条边有美观度和长度。

Ryoku 会使用这样一个策略探索世界：在每个点寻找一个端点她未走过的边中美观度最高的走，如果没有边走，就沿着她前往这个点的边返回，类似于图的深度优先遍历。

探索的一个方案的长度是这个方案所经过的所有边长度的和（返回时经过的长度不用计算）。

她想知道，对于每一个起点 $s=1,2,\cdots ,n$，她需要走过的长度是多少？

输入格式

输入包含 $n+1$ 行，其中第一行包含一个整数 $n$。
接下来 $n$ 行每行包含四个整数 $u,v,w,p$，描述了一条连接 $u$ 和 $v$，长度为 $w$，美观度为 $p$ 的无向边。

输出格式

输出包含 $n$ 行，每行一个整数，第 $i$ 行为 $s=i$ 时的答案。

输入输出样例 #1

输入 #1

5
4 1 2 1
1 2 3 2
3 1 1 4
3 5 2 5
2 3 2 3

输出 #1

7
7
8
7
8

说明/提示

【样例 1 说明】

以下为输入输出样例 1 中的图： （边上红色数组为 $p$，黑色为 $w$）

若起点为 $1$，顺序为 $1\to 3\to 5\to 2\to 4$，长度之和为 $7$。
若起点为 $2$，顺序为 $2\to 3\to 5\to 1\to 4$，长度之和为 $7$。
若起点为 $3$，顺序为 $3\to 5\to 1\to 2\to 4$，长度之和为 $8$。
若起点为 $4$，顺序为 $4\to 1\to 3\to 5\to 2$，长度之和为 $7$。
若起点为 $5$，顺序为 $5\to 3\to 1\to 2\to 4$，长度之和为 $8$。

---

【数据规模与约定】

对于 $40\%$ 的数据，$n\le 10^3$。
对于 $100\%$ 的数据，$3\le n\le 10^6$，$1\le u,v,p\le n$，$0\le w\le 10^9$，保证 $p$ 互不相同。

P6037 Ryoku 的探索

性质一：n点n条边，无向连通意味着有且只有一个环。任意点为起点，经过n条边。如果没有重复顶点，则总过有n+1个连通顶点；如果有2个重复顶点，则只能连通n-1个顶点。
如果s是环上一点，环上相邻两点分别为i1,i2。不失一般性，假定i1的美丽度低，则除边s到i1外，其它边都会访问。
如果s不是环上一点，s1是距离s最近的环上点。则结果和s1同。
一，拓扑排序计算各点是否是环上一点，用ring[i]记录。
二，neiBo1记录，环上个边。记录三个信息：临接点，美丽度、权。有且只有两个点，如果neiBo1[i][0]的美丽大于neiBo1[i][1]后，交换之。
三，neiBo2记录环上各边，只需要记录临接点。
四，a[i]记录距离i最近的环上点。以环上各点为源，BFS各点，BFS状态包括：起点，当前点。
五，所有边权之和- beiBo1[a[s]][0]的边权。

代码

核心代码

#include <iostream>
#include <sstream>
#include <vector>
#include<map>
#include<unordered_map>
#include<set>
#include<unordered_set>
#include<string>
#include<algorithm>
#include<functional>
#include<queue>
#include <stack>
#include<iomanip>
#include<numeric>
#include <math.h>
#include <climits>
#include<assert.h>
#include<cstring>
#include<list>#include <bitset>
using namespace std;template<class T1, class T2>
std::istream& operator >> (std::istream& in, pair<T1, T2>& pr) {in >> pr.first >> pr.second;return in;
}template<class T1, class T2, class T3 >
std::istream& operator >> (std::istream& in, tuple<T1, T2, T3>& t) {in >> get<0>(t) >> get<1>(t) >> get<2>(t);return in;
}template<class T1, class T2, class T3, class T4 >
std::istream& operator >> (std::istream& in, tuple<T1, T2, T3, T4>& t) {in >> get<0>(t) >> get<1>(t) >> get<2>(t) >> get<3>(t);return in;
}template<class T = int>
vector<T> Read() {int n;cin >> n;vector<T> ret(n);for (int i = 0; i < n; i++) {cin >> ret[i];}return ret;
}
template<class T = int>
vector<T> ReadNotNum() {vector<T> ret;T tmp;while (cin >> tmp){ret.emplace_back(tmp);if ('\n' == cin.get()) { break; }} return ret;
}template<class T = int>
vector<T> Read(int n) {vector<T> ret(n);for (int i = 0; i < n; i++) {cin >> ret[i];}return ret;
}template<int N = 1'000'000>
class COutBuff
{
public:COutBuff() {m_p = puffer;}template<class T>void write(T x) {int num[28], sp = 0;if (x < 0)*m_p++ = '-', x = -x;if (!x)*m_p++ = 48;while (x)num[++sp] = x % 10, x /= 10;while (sp)*m_p++ = num[sp--] + 48;AuotToFile();}void writestr(const char* sz) {strcpy(m_p, sz);m_p += strlen(sz);AuotToFile();}inline void write(char ch){*m_p++ = ch;AuotToFile();}inline void ToFile() {fwrite(puffer, 1, m_p - puffer, stdout);m_p = puffer;}~COutBuff() {ToFile();}
private:inline void AuotToFile() {if (m_p - puffer > N - 100) {ToFile();}}char  puffer[N], * m_p;
};template<int N = 1'000'000>
class CInBuff
{
public:inline CInBuff() {}inline CInBuff<N>& operator>>(char& ch) {FileToBuf();ch = *S++;return *this;}inline CInBuff<N>& operator>>(int& val) {FileToBuf();int x(0), f(0);while (!isdigit(*S))f |= (*S++ == '-');while (isdigit(*S))x = (x << 1) + (x << 3) + (*S++ ^ 48);val = f ? -x : x; S++;//忽略空格换行		return *this;}inline CInBuff& operator>>(long long& val) {FileToBuf();long long x(0); int f(0);while (!isdigit(*S))f |= (*S++ == '-');while (isdigit(*S))x = (x << 1) + (x << 3) + (*S++ ^ 48);val = f ? -x : x; S++;//忽略空格换行return *this;}template<class T1, class T2>inline CInBuff& operator>>(pair<T1, T2>& val) {*this >> val.first >> val.second;return *this;}template<class T1, class T2, class T3>inline CInBuff& operator>>(tuple<T1, T2, T3>& val) {*this >> get<0>(val) >> get<1>(val) >> get<2>(val);return *this;}template<class T1, class T2, class T3, class T4>inline CInBuff& operator>>(tuple<T1, T2, T3, T4>& val) {*this >> get<0>(val) >> get<1>(val) >> get<2>(val) >> get<3>(val);return *this;}template<class T = int>inline CInBuff& operator>>(vector<T>& val) {int n;*this >> n;val.resize(n);for (int i = 0; i < n; i++) {*this >> val[i];}return *this;}template<class T = int>vector<T> Read(int n) {vector<T> ret(n);for (int i = 0; i < n; i++) {*this >> ret[i];}return ret;}template<class T = int>vector<T> Read() {vector<T> ret;*this >> ret;return ret;}
private:inline void FileToBuf() {const int canRead = m_iWritePos - (S - buffer);if (canRead >= 100) { return; }if (m_bFinish) { return; }for (int i = 0; i < canRead; i++){buffer[i] = S[i];//memcpy出错			}m_iWritePos = canRead;buffer[m_iWritePos] = 0;S = buffer;int readCnt = fread(buffer + m_iWritePos, 1, N - m_iWritePos, stdin);if (readCnt <= 0) { m_bFinish = true; return; }m_iWritePos += readCnt;buffer[m_iWritePos] = 0;S = buffer;}int m_iWritePos = 0; bool m_bFinish = false;char buffer[N + 10], * S = buffer;
};class KMP
{
public:virtual int Find(const string& s, const string& t){CalLen(t);for (int i1 = 0, j = 0; i1 < s.length(); ){for (; (j < t.length()) && (i1 + j < s.length()) && (s[i1 + j] == t[j]); j++);//i2 = i1 + j 此时s[i1,i2)和t[0,j)相等 s[i2]和t[j]不存在或相等//t[0,j)的结尾索引是j-1，所以最长公共前缀为m_vLen[j-1]，简写为y 则t[0,y)等于t[j-y,j)等于s[i2-y,i2)if (0 == j){i1++;continue;}const int i2 = i1 + j;j = m_vLen[j - 1];i1 = i2 - j;//i2不变}return -1;}//vector<int> m_vSameLen;//m_vSame[i]记录 s[i...]和t[0...]最长公共前缀，增加可调试性 部分m_vSameLen[i]会缺失//static vector<int> Next(const string& s)//{// j = vNext[i] 表示s[0,i]的最大公共前后缀是s[0,j]//	const int len = s.length();//	vector<int> vNext(len, -1);//	for (int i = 1; i < len; i++)//	{//		int next = vNext[i - 1];//		while ((-1 != next) && (s[next + 1] != s[i]))//		{//			next = vNext[next];//		}//		vNext[i] = next + (s[next + 1] == s[i]);//	}//	return vNext;//}const vector<int> CalLen(const string& str){m_vLen.resize(str.length());for (int i = 1; i < str.length(); i++){int next = m_vLen[i - 1];while (str[next] != str[i]){if (0 == next){break;}next = m_vLen[next - 1];}m_vLen[i] = next + (str[next] == str[i]);}return m_vLen;}
protected:int m_c;vector<int> m_vLen;//m_vLen[i] 表示str[0,i]的最长公共前后缀的长度
};class CUnionFind
{
public:CUnionFind(int iSize) :m_vNodeToRegion(iSize){for (int i = 0; i < iSize; i++){m_vNodeToRegion[i] = i;}m_iConnetRegionCount = iSize;}CUnionFind(vector<vector<int>>& vNeiBo) :CUnionFind(vNeiBo.size()){for (int i = 0; i < vNeiBo.size(); i++) {for (const auto& n : vNeiBo[i]) {Union(i, n);}}}int GetConnectRegionIndex(int iNode){int& iConnectNO = m_vNodeToRegion[iNode];if (iNode == iConnectNO){return iNode;}return iConnectNO = GetConnectRegionIndex(iConnectNO);}void Union(int iNode1, int iNode2){const int iConnectNO1 = GetConnectRegionIndex(iNode1);const int iConnectNO2 = GetConnectRegionIndex(iNode2);if (iConnectNO1 == iConnectNO2){return;}m_iConnetRegionCount--;if (iConnectNO1 > iConnectNO2){UnionConnect(iConnectNO1, iConnectNO2);}else{UnionConnect(iConnectNO2, iConnectNO1);}}bool IsConnect(int iNode1, int iNode2){return GetConnectRegionIndex(iNode1) == GetConnectRegionIndex(iNode2);}int GetConnetRegionCount()const{return m_iConnetRegionCount;}vector<int> GetNodeCountOfRegion()//各联通区域的节点数量{const int iNodeSize = m_vNodeToRegion.size();vector<int> vRet(iNodeSize);for (int i = 0; i < iNodeSize; i++){vRet[GetConnectRegionIndex(i)]++;}return vRet;}std::unordered_map<int, vector<int>> GetNodeOfRegion(){std::unordered_map<int, vector<int>> ret;const int iNodeSize = m_vNodeToRegion.size();for (int i = 0; i < iNodeSize; i++){ret[GetConnectRegionIndex(i)].emplace_back(i);}return ret;}
private:void UnionConnect(int iFrom, int iTo){m_vNodeToRegion[iFrom] = iTo;}vector<int> m_vNodeToRegion;//各点所在联通区域的索引,本联通区域任意一点的索引，为了增加可理解性，用最小索引int m_iConnetRegionCount;
};class CTopSort
{
public:template <class T = vector<int> >void Init(const vector<T>& vNeiBo, bool bDirect){const int iDelOutDeg = bDirect ? 0 : 1;m_c = vNeiBo.size();m_vBackNeiBo.resize(m_c);vector<int> vOutDeg(m_c);for (int cur = 0; cur < m_c; cur++){vOutDeg[cur] = vNeiBo[cur].size();for (const auto& next : vNeiBo[cur]){m_vBackNeiBo[next].emplace_back(cur);}}vector<bool> m_vHasDo(m_c);queue<int> que;for (int i = 0; i < m_c; i++){if (vOutDeg[i] <= iDelOutDeg){m_vHasDo[i] = true;if (OnDo(i)) {que.emplace(i);}}}while (que.size()){const int cur = que.front();que.pop();for (const auto& next : m_vBackNeiBo[cur]){if (m_vHasDo[next]){continue;}vOutDeg[next]--;if (vOutDeg[next] <= iDelOutDeg){m_vHasDo[next] = true;if (OnDo(next)) {que.emplace(next);}}}};}int m_c;
protected:virtual bool OnDo(int cur) = 0;vector<vector<int>> m_vBackNeiBo;};
class CMyTopSort :public CTopSort
{
public:CMyTopSort(int N) :m_ring(N, true) {}virtual bool OnDo(int cur) override{m_ring[cur] = false;return true;}vector<bool> m_ring;
};class Solution {
public:vector<long long> Ans(const int N, vector<tuple<int, int, int, int>>& uvwp) {vector<vector<int>> neiBo(N);for (auto& [u, v, w, p] : uvwp) {u--, v--;neiBo[u].emplace_back(v);neiBo[v].emplace_back(u);}CMyTopSort topSort(N);topSort.Init(neiBo, false);vector<vector<tuple<int, int, int>>> neiBo1(N);vector<vector<int>> neiBo2(N);long long total = 0;for (const auto& [u, v, w, p] : uvwp) {total += w;if (topSort.m_ring[u] && topSort.m_ring[v]) {neiBo1[u].emplace_back(v, w, p);neiBo1[v].emplace_back(u, w, p);}else {neiBo2[u].emplace_back(v);neiBo2[v].emplace_back(u);}}for (auto& v : neiBo1) {if (v.size() && (get<2>(v[0]) > get<2>(v[1]))) {swap(v[0], v[1]);}}vector<int> a(N, -1);queue<pair<int, int>> que;auto Add = [&](int pa, int cur) {if (-1 != a[cur]) { return; }que.emplace(pa, cur);a[cur] = pa;};for (int i = 0; i < N; i++) {if (topSort.m_ring[i]) { Add(i, i); }}while (que.size()) {auto [pa, cur] = que.front(); que.pop();for (const auto& next : neiBo2[cur]) {Add(pa, next);}}vector<long long> ans;for (int i = 0; i < N; i++) {ans.emplace_back(total - get<1>(neiBo1[a[i]][0]));}return ans;}
};int main() {
#ifdef _DEBUGfreopen("a.in", "r", stdin);
#endif // DEBUG	ios::sync_with_stdio(0);int n; cin >> n ;auto edge = Read<tuple<int, int,int,int>>(n);
#ifdef _DEBUG		printf("N=%d", n);	Out(edge, ",edge=");	//Out(edge2, ",edge2=");/*Out(que, "que=");*/
#endif // DEBUG	auto res = Solution().Ans(n,edge);for (const auto& i : res){cout << i << "\n";}return 0;
}

单元测试

int N;vector<tuple<int, int, int, int>> edge;TEST_METHOD(TestMethod11){N = 5, edge = { {4,1,2,1},{1,2,3,2},{3,1,1,4},{3,5,2,5},{2,3,2,3} };auto res = Solution().Ans(N,edge);AssertEx({ 7,7,8,7,8 }, res);}

内存超限的解决方案

拓扑排序改成无向图，不需要反向邻接表。优化掉一个临接表。1000‘000个空向量大约需要20M内存，本题才125M。所以要减少临接表数量。

代码

#include <iostream>
#include <sstream>
#include <vector>
#include<map>
#include<unordered_map>
#include<set>
#include<unordered_set>
#include<string>
#include<algorithm>
#include<functional>
#include<queue>
#include <stack>
#include<iomanip>
#include<numeric>
#include <math.h>
#include <climits>
#include<assert.h>
#include<cstring>
#include<list>#include <bitset>
using namespace std;template<class T1, class T2>
std::istream& operator >> (std::istream& in, pair<T1, T2>& pr) {in >> pr.first >> pr.second;return in;
}template<class T1, class T2, class T3 >
std::istream& operator >> (std::istream& in, tuple<T1, T2, T3>& t) {in >> get<0>(t) >> get<1>(t) >> get<2>(t);return in;
}template<class T1, class T2, class T3, class T4 >
std::istream& operator >> (std::istream& in, tuple<T1, T2, T3, T4>& t) {in >> get<0>(t) >> get<1>(t) >> get<2>(t) >> get<3>(t);return in;
}template<class T = int>
vector<T> Read() {int n;cin >> n;vector<T> ret(n);for (int i = 0; i < n; i++) {cin >> ret[i];}return ret;
}
template<class T = int>
vector<T> ReadNotNum() {vector<T> ret;T tmp;while (cin >> tmp) {ret.emplace_back(tmp);if ('\n' == cin.get()) { break; }}return ret;
}template<class T = int>
vector<T> Read(int n) {vector<T> ret(n);for (int i = 0; i < n; i++) {cin >> ret[i];}return ret;
}template<int N = 1'000'000>
class COutBuff
{
public:COutBuff() {m_p = puffer;}template<class T>void write(T x) {int num[28], sp = 0;if (x < 0)*m_p++ = '-', x = -x;if (!x)*m_p++ = 48;while (x)num[++sp] = x % 10, x /= 10;while (sp)*m_p++ = num[sp--] + 48;AuotToFile();}void writestr(const char* sz) {strcpy(m_p, sz);m_p += strlen(sz);AuotToFile();}inline void write(char ch){*m_p++ = ch;AuotToFile();}inline void ToFile() {fwrite(puffer, 1, m_p - puffer, stdout);m_p = puffer;}~COutBuff() {ToFile();}
private:inline void AuotToFile() {if (m_p - puffer > N - 100) {ToFile();}}char  puffer[N], * m_p;
};template<int N = 1'000'000>
class CInBuff
{
public:inline CInBuff() {}inline CInBuff<N>& operator>>(char& ch) {FileToBuf();ch = *S++;return *this;}inline CInBuff<N>& operator>>(int& val) {FileToBuf();int x(0), f(0);while (!isdigit(*S))f |= (*S++ == '-');while (isdigit(*S))x = (x << 1) + (x << 3) + (*S++ ^ 48);val = f ? -x : x; S++;//忽略空格换行		return *this;}inline CInBuff& operator>>(long long& val) {FileToBuf();long long x(0); int f(0);while (!isdigit(*S))f |= (*S++ == '-');while (isdigit(*S))x = (x << 1) + (x << 3) + (*S++ ^ 48);val = f ? -x : x; S++;//忽略空格换行return *this;}template<class T1, class T2>inline CInBuff& operator>>(pair<T1, T2>& val) {*this >> val.first >> val.second;return *this;}template<class T1, class T2, class T3>inline CInBuff& operator>>(tuple<T1, T2, T3>& val) {*this >> get<0>(val) >> get<1>(val) >> get<2>(val);return *this;}template<class T1, class T2, class T3, class T4>inline CInBuff& operator>>(tuple<T1, T2, T3, T4>& val) {*this >> get<0>(val) >> get<1>(val) >> get<2>(val) >> get<3>(val);return *this;}template<class T = int>inline CInBuff& operator>>(vector<T>& val) {int n;*this >> n;val.resize(n);for (int i = 0; i < n; i++) {*this >> val[i];}return *this;}template<class T = int>vector<T> Read(int n) {vector<T> ret(n);for (int i = 0; i < n; i++) {*this >> ret[i];}return ret;}template<class T = int>vector<T> Read() {vector<T> ret;*this >> ret;return ret;}
private:inline void FileToBuf() {const int canRead = m_iWritePos - (S - buffer);if (canRead >= 100) { return; }if (m_bFinish) { return; }for (int i = 0; i < canRead; i++){buffer[i] = S[i];//memcpy出错			}m_iWritePos = canRead;buffer[m_iWritePos] = 0;S = buffer;int readCnt = fread(buffer + m_iWritePos, 1, N - m_iWritePos, stdin);if (readCnt <= 0) { m_bFinish = true; return; }m_iWritePos += readCnt;buffer[m_iWritePos] = 0;S = buffer;}int m_iWritePos = 0; bool m_bFinish = false;char buffer[N + 10], * S = buffer;
};class KMP
{
public:virtual int Find(const string& s, const string& t){CalLen(t);for (int i1 = 0, j = 0; i1 < s.length(); ){for (; (j < t.length()) && (i1 + j < s.length()) && (s[i1 + j] == t[j]); j++);//i2 = i1 + j 此时s[i1,i2)和t[0,j)相等 s[i2]和t[j]不存在或相等//t[0,j)的结尾索引是j-1，所以最长公共前缀为m_vLen[j-1]，简写为y 则t[0,y)等于t[j-y,j)等于s[i2-y,i2)if (0 == j){i1++;continue;}const int i2 = i1 + j;j = m_vLen[j - 1];i1 = i2 - j;//i2不变}return -1;}//vector<int> m_vSameLen;//m_vSame[i]记录 s[i...]和t[0...]最长公共前缀，增加可调试性 部分m_vSameLen[i]会缺失//static vector<int> Next(const string& s)//{// j = vNext[i] 表示s[0,i]的最大公共前后缀是s[0,j]//	const int len = s.length();//	vector<int> vNext(len, -1);//	for (int i = 1; i < len; i++)//	{//		int next = vNext[i - 1];//		while ((-1 != next) && (s[next + 1] != s[i]))//		{//			next = vNext[next];//		}//		vNext[i] = next + (s[next + 1] == s[i]);//	}//	return vNext;//}const vector<int> CalLen(const string& str){m_vLen.resize(str.length());for (int i = 1; i < str.length(); i++){int next = m_vLen[i - 1];while (str[next] != str[i]){if (0 == next){break;}next = m_vLen[next - 1];}m_vLen[i] = next + (str[next] == str[i]);}return m_vLen;}
protected:int m_c;vector<int> m_vLen;//m_vLen[i] 表示str[0,i]的最长公共前后缀的长度
};class CUnionFind
{
public:CUnionFind(int iSize) :m_vNodeToRegion(iSize){for (int i = 0; i < iSize; i++){m_vNodeToRegion[i] = i;}m_iConnetRegionCount = iSize;}CUnionFind(vector<vector<int>>& vNeiBo) :CUnionFind(vNeiBo.size()){for (int i = 0; i < vNeiBo.size(); i++) {for (const auto& n : vNeiBo[i]) {Union(i, n);}}}int GetConnectRegionIndex(int iNode){int& iConnectNO = m_vNodeToRegion[iNode];if (iNode == iConnectNO){return iNode;}return iConnectNO = GetConnectRegionIndex(iConnectNO);}void Union(int iNode1, int iNode2){const int iConnectNO1 = GetConnectRegionIndex(iNode1);const int iConnectNO2 = GetConnectRegionIndex(iNode2);if (iConnectNO1 == iConnectNO2){return;}m_iConnetRegionCount--;if (iConnectNO1 > iConnectNO2){UnionConnect(iConnectNO1, iConnectNO2);}else{UnionConnect(iConnectNO2, iConnectNO1);}}bool IsConnect(int iNode1, int iNode2){return GetConnectRegionIndex(iNode1) == GetConnectRegionIndex(iNode2);}int GetConnetRegionCount()const{return m_iConnetRegionCount;}vector<int> GetNodeCountOfRegion()//各联通区域的节点数量{const int iNodeSize = m_vNodeToRegion.size();vector<int> vRet(iNodeSize);for (int i = 0; i < iNodeSize; i++){vRet[GetConnectRegionIndex(i)]++;}return vRet;}std::unordered_map<int, vector<int>> GetNodeOfRegion(){std::unordered_map<int, vector<int>> ret;const int iNodeSize = m_vNodeToRegion.size();for (int i = 0; i < iNodeSize; i++){ret[GetConnectRegionIndex(i)].emplace_back(i);}return ret;}
private:void UnionConnect(int iFrom, int iTo){m_vNodeToRegion[iFrom] = iTo;}vector<int> m_vNodeToRegion;//各点所在联通区域的索引,本联通区域任意一点的索引，为了增加可理解性，用最小索引int m_iConnetRegionCount;
};class CUGTopSort
{
public:template <class T = vector<int> >CUGTopSort(const vector<T>& vNeiBo) :m_vDeg(vNeiBo.size()) {const int N = vNeiBo.size();m_vHas.resize(N);for (int u = 0; u < N; u++) {for (const auto& v : vNeiBo[u]) {m_vDeg[u]++;}}queue<int> que;auto Add = [&](int i) {if (m_vDeg[i] <= 1){if (m_vHas[i]) { return; }m_vHas[i] = true;if (OnDo(i)) {que.emplace(i);}}};for (int i = 0; i < N; i++){Add(i);}while (que.size()){const int cur = que.front();que.pop();for (const auto& next : vNeiBo[cur]){m_vDeg[next]--;Add(next);}};}vector<int> m_vDeg;vector<bool> m_vHas;
protected:virtual bool OnDo(int cur) { return true; };
};class Solution {
public:vector<long long> Ans(const int N, vector<tuple<int, int, int, int>>& uvwp) {vector<vector<int>> neiBo(N);long long total = 0;for (auto& [u, v, w, p] : uvwp) {total += w;u--, v--;neiBo[u].emplace_back(v);neiBo[v].emplace_back(u);}CUGTopSort topSort(neiBo);vector<vector<tuple<int, int, int>>> neiBo1(N);for (const auto& [u, v, w, p] : uvwp) {if ((topSort.m_vDeg[u] > 1) && (topSort.m_vDeg[v] > 1)) {neiBo1[u].emplace_back(v, w, p);neiBo1[v].emplace_back(u, w, p);}}for (auto& v : neiBo1) {if (v.size() && (get<2>(v[0]) > get<2>(v[1]))) {swap(v[0], v[1]);}}vector<int> a(N, -1);queue<pair<int, int>> que;auto Add = [&](int pa, int cur) {if (-1 != a[cur]) { return; }que.emplace(pa, cur);a[cur] = pa;};for (int i = 0; i < N; i++) {if (topSort.m_vDeg[i] > 1) { Add(i, i); }}while (que.size()) {auto [pa, cur] = que.front(); que.pop();for (const auto& next : neiBo[cur]) {Add(pa, next);}}vector<long long> ans;for (int i = 0; i < N; i++) {ans.emplace_back(total - get<1>(neiBo1[a[i]][0]));}return ans;}
};int main() {
#ifdef _DEBUGfreopen("a.in", "r", stdin);
#endif // DEBUG	ios::sync_with_stdio(0);int n;cin >> n;auto edge = Read<tuple<int, int, int, int>>(n);
#ifdef _DEBUG		printf("N=%d", n);Out(edge, ",edge=");//Out(edge2, ",edge2=");/*Out(que, "que=");*/
#endif // DEBUG	auto res = Solution().Ans(n, edge);for (const auto& i : res){cout << i << "\n";}return 0;
}

扩展阅读

我想对大家说的话
工作中遇到的问题，可以按类别查阅鄙人的算法文章，请点击《算法与数据汇总》。
学习算法：按章节学习《喜缺全书算法册》，大量的题目和测试用例，打包下载。重视操作
有效学习：明确的目标 及时的反馈 拉伸区（难度合适） 专注
闻缺陷则喜(喜缺)是一个美好的愿望，早发现问题，早修改问题，给老板节约钱。
子墨子言之：事无终始，无务多业。也就是我们常说的专业的人做专业的事。
如果程序是一条龙，那算法就是他的是睛
失败+反思=成功 成功+反思=成功

视频课程

先学简单的课程，请移步CSDN学院，听白银讲师（也就是鄙人）的讲解。
https://edu.csdn.net/course/detail/38771
如何你想快速形成战斗了，为老板分忧，请学习C#入职培训、C++入职培训等课程
https://edu.csdn.net/lecturer/6176

测试环境

操作系统：win7 开发环境： VS2019 C++17
或者 操作系统：win10 开发环境： VS2022 C++17
如无特殊说明，本算法用**C++**实现。