POJ 3259 Wormholes(最短路)

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

题目大意
判断 John 是否能实现从某一处出发，然后返回原地时间在出发时间之前，在道路中存在虫洞，穿过虫洞时间倒流且虫洞是单向的，而其余路径是双向的。

解题思路
边的权值即为经过此通路的时间，使用Bellman-Ford 算法对边进行松弛，只需判断是否存在负环。
注意输入：虫洞是负权，其余通路是正权。

代码实现

#include <iostream>
#include<cstdio>
using namespace std;
#define maxn 520
#define maxm 6006
const int INF=(1<<26);
struct node
{
    int s,e,w;
}edge[maxm];
int dis[maxn],countt;
bool bellman_ford(int n)
{
    int s,e,w;
    for(int i=1;i<=n;i++)
        dis[i]=INF;
    dis[1]=0;
    for(int i=0;i<n;i++)
        for(int j=0;j<countt;j++)
    {
        s=edge[j].s;
        e=edge[j].e;
        w=edge[j].w;
        if(dis[s]<INF&&dis[e]>dis[s]+w)
        {
            dis[e]=dis[s]+w;
            if(i==n-1)
                return true;
        }
    }
    return false;
}
int main()
{
    int T,n,m,l,s,e,w;
    cin>>T;
    while(T--)
    {
        countt=0;
        cin>>n>>m>>l;
        for(int i=0;i<m;i++)
        {
            cin>>s>>e>>w;
            edge[countt].s=edge[countt+1].e=s;
            edge[countt+1].s=edge[countt].e=e;
            edge[countt++].w=w;
            edge[countt++].w=w;
        }
        for(int i=0;i<l;i++)
        {
            cin>>s>>e>>w;
            edge[countt].s=s;
            edge[countt].e=e;
            edge[countt++].w=-w;
        }
        if(bellman_ford(n)) printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}

#include <iostream>
#include<cstdio>
#include<cstring>
#include<stdlib.h>
using namespace std;
#define maxn 2507*2+200
#define maxv 507
struct edge
{
    int from;
    int to;
    int cost;
} ed[maxn];
int d[maxv],v,e;
bool find_negative_loop()
{
    memset(d,0,sizeof(d));
    for(int i=0; i<v; i++)
    {
        for(int j=0; j<e; j++)
        {
            edge es=ed[j];
            if(d[es.to]>d[es.from]+es.cost)
            {
                d[es.to]=d[es.from]+es.cost;
                if(i==v-1) return true;
            }
        }
    }
    return false;
}

int main()
{
    int T;
    int m,w;
    int f,t,c;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d %d %d",&v,&m,&w);
        e=0;
        for(int i=0; i<m; i++)
        {
            scanf("%d %d %d",&f,&t,&c);
            ed[e].from=f;
            ed[e].to=t;
            ed[e].cost=c;
            e++;
            ed[e].from=t;
            ed[e].to=f;
            ed[e].cost=c;
            e++;
        }
        for(int i=m; i<m+w; i++)
        {
            scanf("%d %d %d",&ed[e].from,&ed[e].to,&c);
            ed[e].cost=-c;
            e++;
        }
        if(find_negative_loop())
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}