poj 3259 Wormholes

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

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

Sample Output

NOYES

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.

题意分析：大体意思就是有f个农场，有n个点，有m条双向路，有w个单向虫洞连接着个点，构造最短路径，看是否存在负环，就是通过虫洞倒退时间看能够存在负数环，来使一个点的出发时间小于到达时间，用bellman就可以过

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#include<queue>using namespace std;#define ll long longconst int maxn=5202;struct node{    int u,v,w;};node edge[maxn]; int n,m,p,dist[maxn],k;void addedge(int u,int v,int w){    edge[k].u=u;    edge[k].v=v;    edge[k].w=w;    k++;}bool bellman_ford(){    int i,j;    memset(dist,0,sizeof(dist));    for (i=1;i<=n;i++)        for (j=0;j<k;j++)            if (dist[edge[j].u]+edge[j].w<dist[edge[j].v])                 dist[edge[j].v]=dist[edge[j].u]+edge[j].w;    for (j=0;j<k;j++)         if (dist[edge[j].u]+edge[j].w<dist[edge[j].v])            return 0;    return 1;}int main(){    int t;    int i,u,v,w;    scanf("%d",&t);    while (t--)    {        k=0;        scanf("%d%d%d",&n,&m,&p);        for (i=0;i<m;i++)        {            scanf("%d%d%d",&u,&v,&w);            addedge(u,v,w);            addedge(v,u,w);        }        for (i=0;i<p;i++)        {            scanf("%d%d%d",&u,&v,&w);            addedge(u,v,-w);        }        if (bellman_ford()==0)            printf("YES\n");        else            printf("NO\n");    }    return 0;}

spfa算法一样

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#include<queue>using namespace std;#define ll long longconst int INF=1<<30;const int maxn=5202;struct node{    int v,w,next;};node edge[maxn];int m,n,p,k,dist[maxn],head[maxn],cnt[maxn];bool vis[maxn];void addedge(int u,int v,int w){    edge[k].v=v;    edge[k].w=w;    edge[k].next=head[u];    head[u]=k++;}bool spfa(){    int i;    memset(vis,0,sizeof(vis));    memset(cnt,0,sizeof(cnt));    for (i=0;i<k;i++)        dist[i]=INF;    cnt[1]++;    dist[1]=0;    vis[1]=1;    queue<int>q;    q.push(1);    while (!q.empty())    {        int x=q.front();        q.pop();        vis[x]=0;        for (int e=head[x];e!=-1;e=edge[e].next)        {            if (dist[edge[e].v]>dist[x]+edge[e].w)            {                dist[edge[e].v]=dist[x]+edge[e].w;                if (!vis[edge[e].v])                {                    q.push(edge[e].v);                    vis[edge[e].v]=1;                    cnt[edge[e].v]++;                    if (cnt[edge[e].v]>=n)                        return 0;                }            }        }    }    return 1;}int main(){   int t,i,j,u,v,w;   scanf("%d",&t);   while (t--)   {       k=0;     scanf("%d%d%d",&n,&m,&p);     memset(head,-1,sizeof(head));       for (i=0;i<m;i++)       {           scanf("%d%d%d",&u,&v,&w);           addedge(u,v,w);           addedge(v,u,w);       }       for (i=0;i<p;i++)       {           scanf("%d%d%d",&u,&v,&w);           addedge(u,v,-w);       }       if (spfa())        printf("NO\n");       else        printf("YES\n");   }    return 0;}