POJ3259 :Wormholes(SPFA判负环）

POJ3259 :Wormholes

时间限制:2000MS 内存限制:65536KByte 64位IO格式:%I64d & %I64u

描述

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.

输入

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.

输出

Lines 1..F: For each farm, output "YES" if FJ can achieve his goal, otherwise output "NO" (do not include the quotes).

样例输入

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

样例输出

NO
YES

提示

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：

FJ的农场有很多虫洞，可以实现时光倒流，现在给你农场每条道路的起点、终点和时间，还有虫洞的起点、终点和能倒流的
时间，问FJ能不能通过时光倒流看到之前的自己。
首先T组数据，再输入n，m，w 分别代表n个节点，m个无向边的信息，w个有向虫洞的信息

思路：

把w个虫洞当负值加进边就行

比如第一组数据的3 1 3，在构图的时候add_edge(3,1,-3)，即可。

其实就是用spfa跑一遍图，看一下有没有负环即可。有的话输出YES，没有的话输出NO

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
#include<cmath>
#include<vector>
#include<cstring>
using namespace std;
typedef long long LL;
#define Max_N 1001
#define INF 100000000
int head[Max_N],vis[Max_N],n,In[Max_N],cnt;
int dis[Max_N];
int pa[Max_N];
struct note{
    int from,to,next;
    int c;
}edge[50005*2];
void init(){
    memset(head,-1,sizeof(head));
    cnt = 0;
}
void add_edge(int u,int v,double c){
    edge[cnt].from = u;
    edge[cnt].to = v;
    edge[cnt].c = c;
    edge[cnt].next = head[u];
    head[u] = cnt++;
}
bool spfa(int s,int n,int key){
    queue<int>q;
    for(int i = 0 ;i <= n ; i++)
        dis[i] = INF;
    memset(vis,0,sizeof(vis));
    memset(In,0,sizeof(In));
    q.push(s);
    vis[s] = 1;
    dis[s] = 0;
    if(key == -1)
        memset(pa,-1,sizeof(pa));
    while(!q.empty()){
        int p = q.front();
        vis[p] = 0;
        q.pop();
        for(int i = head[p] ; ~i ; i = edge[i].next){
            if(key==i)continue;
            int temp = edge[i].to;
            if(dis[temp] > dis[p] + edge[i].c){
                dis[temp] = dis[p] + edge[i].c;
                if(key==-1)
                     pa[temp] = i;
                if(!vis[temp]){
                    q.push(temp);
                    vis[temp] = 1;
                    if(++In[temp]>n)return false;
                }
            }
        }
    }
    return true;
}//spfa带记录路径,通过Key调整
int main(){
    int m,t,w;
    for(scanf("%d",&t);t--;){
        scanf("%d %d %d",&n,&m,&w);
        init();
        int x,y,c;
        for(int i = 0 ; i < m ; i++){
            scanf("%d %d %d",&x,&y,&c);
            add_edge(x,y,c);
            add_edge(y,x,c);
        }
        for(int i = 0 ; i < w ; i++){
            scanf("%d %d %d",&x,&y,&c);
            add_edge(x,y,-c);
        }
        spfa(1,n,-1)?puts("NO"):puts("YES");
    }
    return 0;
}