标签：算法 最短路 poj
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 36889 | Accepted: 13520 |
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
Output
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：就是给N组测试数据，给你N个点M条无向边，再给你W个虫洞（可以使时间倒流），问你能不能看到刚出发时的自己。（就是判断存不存在负环）。
思路：套上SPFA公式然后判断负环就可以了。
#include<stdio.h>//好坑，wa了半天没过去，居然是因为大小写问题。整个人都不好了。
#include<string.h>
#include<queue>
#define INF 0x3f3f3f3f
#define M 3000*2
#define N 550
using namespace std;
int n,m,w,dis[N],vis[N],used[N],head[N],edgenum;
struct node{
int from,to,cost,next;
}edge[M];
void init(){
edgenum=0;
memset(head,-1,sizeof(head));
}
void add(int u,int v,int cost){
node E={u,v,cost,head[u]};
edge[edgenum]=E;
head[u]=edgenum++;
}
void spfa(){
queue<int>q;
memset(vis,0,sizeof(vis));
memset(dis,INF,sizeof(dis));
memset(used,0,sizeof(used));
dis[1]=0;
vis[1]=1;
q.push(1);
used[1]++;
while(!q.empty()){
int u=q.front();
q.pop();
vis[u]=0;
for(int i=head[u];i!=-1;i=edge[i].next){
int v=edge[i].to;
if(dis[v]>dis[u]+edge[i].cost){
dis[v]=dis[u]+edge[i].cost;
if(!vis[v]){
vis[v]=1;
used[v]++;
if(used[v]>n){
printf("YES\n");//存在负环就意味着时光可以倒流。
return ;
}
q.push(v);
}
}
}
}
printf("NO\n");
}
int main(){
int T;
scanf("%d",&T);
while(T--){
scanf("%d%d%d",&n,&m,&w);
init();
while(m--){
int a,b,cost;
scanf("%d%d%d",&a,&b,&cost);//添加无向边。
add(a,b,cost);
add(b,a,cost);
}
while(w--){
int a,b,cost;//添加单向虫洞。
scanf("%d%d%d",&a,&b,&cost);
cost=-cost;
add(a,b,cost);
}
spfa();
}
return 0;
}ac代码：
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标签：算法 最短路 poj
原文地址：http://blog.csdn.net/gui951753/article/details/47834083
