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.
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.
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 8Sample Output
NO YESHint
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：每组数据先给出m条正权边再给出w条负权边，其中正权边是双向的而负权边是单向的，判断是否存在负权回路。
解题思路：Bellman-Ford解决负权边。
其原理是：对于n个点的图来说，最多只能进行n-1轮松弛就能找到最短路径，如果在进行n-1轮松弛后仍能继续成功松弛，那么此图必然存在负权回路。
Bellman-Ford算法经常会在未达到n-1轮松弛前就已经计算出最短路。所以添加check用来标记数组dis在本轮松弛中是否发生了变化，如果没有发生变化说明松弛结束，提前跳出循环。
AC代码：
#include<stdio.h>
#define INF 0x3f3f3f3f
int dis[510],n,m,w;
struct edge
{
int s,e,t;
} e[2710];
void bellman()
{
int flag,check;
for(int i=1;i<=n;i++)
dis[i]=INF;
for(int k=1;k<=n-1;k++)
{
check=0;
for(int i=1;i<=m;i++)//双向路
{
if(dis[e[i].s]+e[i].t<dis[e[i].e])
{
dis[e[i].e]=dis[e[i].s]+e[i].t;
check=1;
}
if(dis[e[i].e]+e[i].t<dis[e[i].s])
{
dis[e[i].s]=dis[e[i].e]+e[i].t;
check=1;
}
}
for(int i=m+1;i<=m+w;i++)//单向路
{
if(dis[e[i].s]+e[i].t<dis[e[i].e])
{
dis[e[i].e]=dis[e[i].s]+e[i].t;
check=1;
}
}
if(check==0) break;
}
flag=0;
for(int i=1;i<=m;i++)//判断是否存在负权回路
{
if(dis[e[i].s]+e[i].t<dis[e[i].e]) flag=1;
}
if(flag) printf("YES\n");
else printf("NO\n");
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d%d",&n,&m,&w);
for(int i=1;i<=m;i++)
{
scanf("%d%d%d",&e[i].s,&e[i].e,&e[i].t);
}
for(int i=m+1;i<=m+w;i++)
{
scanf("%d%d%d",&e[i].s,&e[i].e,&e[i].t);
e[i].t=-e[i].t;//转换为负数表示负权
}
bellman();
}
return 0;
}