poj_2240 Bellman-ford算法变形

来源：互联网发布：linux 用终端打开文件编辑：程序博客网时间：2024/06/04 19:46

还是自己尝试了一下bellmanford，变形之处在于要拿每一个点看是否构成正权回路，即每一个点要用一次bellman-ford()。

#include<iostream>#include<fstream>#include<map>using namespace std;int n, m;map<string, int> STL;char str[50], str1[50], str2[50];double dist[40], rate;struct EDGE{int ci;int cj;double cij;}edges[1000];bool bellman_ford(int sr){for (int i = 0; i <= n; i++)dist[i] = 0.0;dist[sr] = 1.0;for (int i = 0; i <n; i++){for (int j = 1; j <= m; j++){if (dist[edges[j].cj] < dist[edges[j].ci] * edges[j].cij){dist[edges[j].cj] = dist[edges[j].ci] * edges[j].cij;}}}if (dist[sr]>1.0)return true;elsereturn false;}int main(){//ifstream in("text.txt");int cases = 0;while (cin >> n && n != 0){cases++;for (int i = 1; i <= n; i++){cin >> str;STL[str] = i;}cin >> m;for (int i = 1; i <= m; i++){cin >> str1 >> rate >> str2;edges[i].ci = STL[str1];edges[i].cj = STL[str2];edges[i].cij = rate;}int sr;for (sr= 1; sr <= n; sr++){if (bellman_ford(sr))break;}if (sr<=n)cout << "Case " << cases << ": Yes" << endl;elsecout << "Case " << cases << ": No" << endl;}//system("pause");return 0;}

0 0