POJ 3259 Wormholes (Bellman-Ford 求负环)（F）

[align=center]Wormholes[/align]

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 51462		Accepted: 19108

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
S and 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 to
E 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

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

Sample Output

NO
YES

Hint
For farm 1, FJ cannot travel back in time.

For farm 2, FJ could travel back in time by the cycle 1->2->3->1, arriving back at his starting location 1 second before he leaves. He could start from anywhere on the cycle to accomplish this.

题意：给出n个点和m条双向边且权值为正，w条单向边，权值为负。要求给定一个图，判断图中是否有负环。

题意：判断是否有环显然需要使用Bellman-Ford算法。因为本题只需判断是否有负环的存在，而不需求最短路，所以可令初始dis均为0，如果第n次松弛成功，则有负环；否则没有。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
#define inf 0x3f3f3f3f
const int maxn=550;

double dis[maxn];

int tot;//边的总数
int way[maxn*maxn][2];//存边的起点和终点
int value[maxn*maxn];//存所花费的时间
int n,m,w;
bool Bellman()
{
for(int i=1;i<=n;i++)
dis[i]=0;
for(int i=0;i<=n;i++)//循环n-1次
{
bool flag=false;//优化
for(int j=0;j<tot;j++)
{
if(dis[way[j][1]]>dis[way[j][0]]+value[j])
{
flag=true;
dis[way[j][1]]=dis[way[j][0]]+value[j];
}
}
if(!flag)return false;//无环
}
return true;
}

int main()
{
int t;
scanf("%d",&t);
int a,b,c;
while(t--)
{
scanf("%d%d%d",&n,&m,&w);
tot=0;
for(int i=0;i<m;i++)
{
scanf("%d%d%d",&a,&b,&c);
way[tot][0]=a;
way[tot][1]=b;
value[tot]=c;
tot++;
way[tot][0]=b;
way[tot][1]=a;
value[tot]=c;
tot++;
}
for(int i=0;i<w;i++)
{
scanf("%d%d%d",&a,&b,&c);
way[tot][0]=a;
way[tot][1]=b;
value[tot]=-c;
tot++;
}
int i;

if(Bellman())
printf("YES\n");
else
printf("NO\n");
}
return 0;
}