标签：des style blog http color os io java strong
Time Limit: 2000 MS Memory Limit: 65536 KB
64-bit integer IO format: %I64d , %I64u Java class name: Main
While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a time that is BEFORE you entered the wormhole! Each of FJ‘s farms comprises N (1 ≤ N ≤ 500) fields conveniently numbered 1..N, M (1 ≤ M ≤ 2500) paths, and W (1 ≤ W ≤ 200) wormholes.
As FJ is an avid time-traveling fan, he wants to do the following: start at some field, travel through some paths and wormholes, and return to the starting field a time before his initial departure. Perhaps he will be able to meet himself :) .
To help FJ find out whether this is possible or not, he will supply you with complete maps to F (1 ≤ F ≤ 5) of his farms. No paths will take longer than 10,000 seconds to travel and no wormhole can bring FJ back in time by more than 10,000 seconds.
2 3 3 1 1 2 2 1 3 4 2 3 1 3 1 3 3 2 1 1 2 3 2 3 4 3 1 8
NO YES
//Memory Time //308K 204MS #include<iostream> #include <string.h> using namespace std; int dis[1001]; //源点到各点权值 const int max_w=10001; //无穷远 struct weight { int s; int e; int t; }edge[5200]; int N,M,W_h; //N (1≤N≤500)fields 顶点数 //M (1≤M≤2500)paths 正权双向边 //W_h (1≤W≤200) wormholes 虫洞（回溯），负权单向边 int all_e; //边集（边总数） bool bellman() { bool flag; /*relax*/ for(int i=0;i<N-1;i++) ///dis松弛的次数 { flag=false; for(int j=0;j<all_e;j++) ///所有边集 if(dis[edge[j].e] > dis[edge[j].s] + edge[j].t) ///可以松弛 更新 { dis[edge[j].e] = dis[edge[j].s] + edge[j].t; flag=true; //relax对路径有更新 } if(!flag) break; //只要某一次relax没有更新，说明最短路径已经查找完毕，或者部分点不可达，可以跳出relax }///已经更新完毕了 所有边集都是两点之间的最短路 /*Search Negative Circle*/ for(int k=0;k<all_e;k++) ///遍历所有边集 如果还出现能够再次更新成最短的边 就表明出现负权值回路了 if( dis[edge[k].e] > dis[edge[k].s] + edge[k].t) return true; return false; } int main(void) { int u,v,w; int F; cin>>F; while(F--) { memset(dis,max_w,sizeof(dis)); //源点到各点的初始值为无穷，即默认不连通 cin>>N>>M>>W_h; all_e=0; //初始化指针 /*read in Positive Paths*/ for(int i=1;i<=M;i++) { cin>>u>>v>>w; edge[all_e].s=edge[all_e+1].e=u; edge[all_e].e=edge[all_e+1].s=v; edge[all_e++].t=w; edge[all_e++].t=w; //由于paths的双向性，两个方向权值相等，注意指针的移动 } /*read in Negative Wormholds*/ for(int j=1;j<=W_h;j++) { cin>>u>>v>>w; edge[all_e].s=u; edge[all_e].e=v; edge[all_e++].t=-w; //注意权值为负 } /*Bellman-Ford Algorithm*/ if(bellman()) cout<<"YES"<<endl; else cout<<"NO"<<endl; } return 0; } //Memory Time //308K 204MS #include<iostream> #include <string.h> using namespace std; int dis[1001]; //源点到各点权值 const int max_w=10001; //无穷远 struct weight { int s; int e; int t; }edge[5200]; int N,M,W_h; //N (1≤N≤500)fields 顶点数 //M (1≤M≤2500)paths 正权双向边 //W_h (1≤W≤200) wormholes 虫洞（回溯），负权单向边 int all_e; //边集（边总数） bool bellman() { bool flag; /*relax*/ for(int i=0;i<N-1;i++) ///dis松弛的次数 { flag=false; for(int j=0;j<all_e;j++) ///所有边集 if(dis[edge[j].e] > dis[edge[j].s] + edge[j].t) ///可以松弛 更新 { dis[edge[j].e] = dis[edge[j].s] + edge[j].t; flag=true; //relax对路径有更新 } if(!flag) break; //只要某一次relax没有更新，说明最短路径已经查找完毕，或者部分点不可达，可以跳出relax }///已经更新完毕了 所有边集都是两点之间的最短路 /*Search Negative Circle*/ for(int k=0;k<all_e;k++) ///遍历所有边集 如果还出现能够再次更新成最短的边 就表明出现负权值回路了 if( dis[edge[k].e] > dis[edge[k].s] + edge[k].t) return true; return false; } int main(void) { int u,v,w; int F; cin>>F; while(F--) { memset(dis,max_w,sizeof(dis)); //源点到各点的初始值为无穷，即默认不连通 cin>>N>>M>>W_h; all_e=0; //初始化指针 /*read in Positive Paths*/ for(int i=1;i<=M;i++) { cin>>u>>v>>w; edge[all_e].s=edge[all_e+1].e=u; edge[all_e].e=edge[all_e+1].s=v; edge[all_e++].t=w; edge[all_e++].t=w; //由于paths的双向性，两个方向权值相等，注意指针的移动 } /*read in Negative Wormholds*/ for(int j=1;j<=W_h;j++) { cin>>u>>v>>w; edge[all_e].s=u; edge[all_e].e=v; edge[all_e++].t=-w; //注意权值为负 } /*Bellman-Ford Algorithm*/ if(bellman()) cout<<"YES"<<endl; else cout<<"NO"<<endl; } return 0; }
标签：des style blog http color os io java strong
原文地址：http://www.cnblogs.com/zhangying/p/3947350.html
