POJ3259：Wormholes（spfa判负环）

Posted 2021-02-07 heyuhhh

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 68097		Accepted: 25374

题目链接：http://poj.org/problem?id=3259

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：

给出一些双向边以及其对应花费，然后还有一些单向边，可以回到之前的时间。问最后能否从起点出发，回到起点的时候时间在出发之前。

 

题解：

对于时间通道建负边权就行了，然后跑spfa看有没得负环。

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int N = 505,M = 2505;
int T,n,m,w;
int head[N],d[N],c[N],vis[N];
struct Edge{
    int u,v,w,next;
}e[M<<2];
int tot;
void adde(int u,int v,int w){
    e[tot].u=u;e[tot].v=v;e[tot].w=w;e[tot].next=head[u];head[u]=tot++;
}
int spfa(int s){
    queue <int> q;
    memset(d,INF,sizeof(d));
    memset(vis,0,sizeof(vis));memset(c,0,sizeof(c));
    q.push(s);vis[s]=1;d[s]=0;c[s]=1;
    while(!q.empty()){
        int u=q.front();q.pop();vis[u]=0;
        if(c[u]>n){
            return 1;
        }
        for(int i=head[u];i!=-1;i=e[i].next){
            int v=e[i].v;
            if(d[v]>d[u]+e[i].w){
                d[v]=d[u]+e[i].w;
                if(!vis[v]){
                    vis[v]=1;
                    q.push(v);
                    c[v]++;
                }
            }
        }
    }
     return 0;
}
int main(){
    scanf("%d",&T);
    while(T--){
        scanf("%d%d%d",&n,&m,&w);
        memset(head,-1,sizeof(head));tot=0;
        for(int i=1;i<=m;i++){
            int u,v,c;
            scanf("%d%d%d",&u,&v,&c);
            adde(u,v,c);adde(v,u,c);
        }
        for(int i=1;i<=w;i++){
            int u,v,c;
            scanf("%d%d%d",&u,&v,&c);
            adde(u,v,-c);
        }
        int flag=0;
        if(spfa(1)) flag=1;
        if(flag) puts("YES");
        else puts("NO");
    }
    return 0;
}