poj_3259_Wormholes(虫洞)

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

题意：
虫洞问题。现在有n个点，m条边，代表现在可以走的通路，比如从a到b和从b到a需要花费c时间，现在在地上出现了w个虫洞，虫洞的意义就是你从a到b话费的时间是-c(时间倒流,并且虫洞是单向的)，现在问你从某个点开始走，能否回到从前。
分析：即检测负权回路，看是否有负权环，有负权环输出YES，否则输出NO；用BellmanFord()算法可以检测是否含有负权回路。

BellmanFord()算法核心语句：

for(k=1;k<=n-1;k++)    for(i=1;i<=m;i++)        if(dis[v[i]] > dis[u[i]]+w[i])            dis[v[i]] = dis[u[i]]+w[i];//检测负权回路int flag=0;for(i=1;i<=m;i++)    if(dis[v[i]] > dis[u[i]]+w[i])        flag=1;if(flag==1)    printf("此图含有负权回路\n");

代码：

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int cnt,m,n,w;int inf = 9999999,dis[5005];struct node{    int u;    int v;    int t;}e[5005];void add(int u,int v,int t){    //因为Bellman-Ford算法要用u，v,t三者之间的关系，所以如果直接存储e[a][b] = c;不好操作    e[cnt].u = u;    e[cnt].v = v;    e[cnt].t = t;    cnt++;}int BellmanFord(){    int i,j;    int flag=0;    for(i=1;i<=n;i++)        dis[i] = inf;    dis[1]=0;    //松弛操作    for(i=0;i<n;i++)    {        for(j=0;j<cnt;j++)        {            if(dis[e[j].v] > dis[e[j].u]+e[j].t)                dis[e[j].v] = dis[e[j].u] + e[j].t;        }    }    //检测负权回路    for(i=0;i<cnt;i++)    {        if(dis[e[i].v] > dis[e[i].u]+e[i].t)            return 1;  //有负权边    }    return flag;   //否则没有}int main(){    int T,i;    int a,b,c;    scanf("%d",&T);    while(T--)    {        scanf("%d%d%d",&n,&m,&w);        cnt=0;        for(i=1;i<=m;i++)        {            scanf("%d%d%d",&a,&b,&c);            add(a,b,c);   //题中说明双向路径            add(b,a,c);        }        for(i=1;i<=w;i++)        {            scanf("%d%d%d",&a,&b,&c);            add(a,b,-c);   //题中已说明单向路径        }        int f = BellmanFord();        if(f==1)            printf("YES\n");        else            printf("NO\n");    }    return 0;}