poj 3259 负权回路+Bellman

题目：

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 45714 Accepted: 16874

Description

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.

Input

Line 1: A single integer, F. F farm descriptions follow. 
Line 1 of each farm: Three space-separated integers respectively: N, M, and W 
Lines 2..M+1 of each farm: Three space-separated numbers (S, E, T) that describe, respectively: a bidirectional path between Sand E that requires T seconds to traverse. Two fields might be connected by more than one path. 
Lines M+2..M+W+1 of each farm: Three space-separated numbers (S, E, T) that describe, respectively: A one way path from S toE that also moves the traveler back T seconds.

Output

Lines 1..F: For each farm, output "YES" if FJ can achieve his goal, otherwise output "NO" (do not include the quotes).

Sample Input

23 3 11 2 21 3 42 3 13 1 33 2 11 2 32 3 43 1 8

Sample Output

NOYES

两组样例，第一组有三个顶点，三条普通双向边，一条可以使时光倒流的特殊边。特殊边读入时需要将第三个值取反。

问图中是否存在负权回路。

分析：

基础bellman。

代码：

#include <stdio.h>#include <iostream>#include <queue>#include <vector>#include <string.h>#include <map> #include <algorithm>#include <math.h>#include <limits.h>using namespace std;const int maxn=510;const int maxe=5300;//2510太小 const int inf=(INT_MAX)/3;struct edge{int from,to,cost;//edge(int from,int to,int cost){//this->from=from;//this->to=to;//this->cost=cost;//}};edge es[maxe];int n,m,w,te;//te表示图中边数 int d[maxn];bool bellfind(){memset(d,0,sizeof(d));for(int i=0;i<n;++i){for(int j=0;j<te;++j){edge e=es[j];if(d[e.to]>d[e.from]+e.cost){d[e.to]=d[e.from]+e.cost;if(i==n-1) return true;}}}return false;}int main(){//720K94MS int t,a,b,c;scanf("%d",&t);while(t--){memset(es,0,sizeof(es));scanf("%d%d%d",&n,&m,&w);for(int i=0;i<m;++i){scanf("%d%d%d",&es[i].from,&es[i].to,&es[i].cost);es[i+m].from=es[i].to;es[i+m].to=es[i].from;es[i+m].cost=es[i].cost;}te=m*2;int c;for(int i=0;i<w;++i){scanf("%d%d%d",&es[i+te].from,&es[i+te].to,&c);es[i+te].cost=-c;}te+=w;if(bellfind()) puts("YES");else puts("NO"); }    return 0;}