poj 3259 判断是否存在负权环
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 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
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
Hint
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.
农夫的农场里存在n块地，存在m条无向路，从每条路的一段到另一端需要花费t的时间，存在w个虫洞，每个虫洞都是单向的从虫洞的起点到虫洞的终点需要-t的时间，也就是时间倒流了。现在农夫要从一块田地出发再回到原出发点，问他能否在回到出发点的时候比刚出发是时间要早。
解题思路：
模型就是一张有负权边的图网络，若能回到原出发点的时候会时间不出发时还少，那就说明一定存在这负权边。这样一来，这道题就变成了判断是否存在负权边了。
在spfa算法中，如果以个顶点入队次数超过了n次，那就说明存在了负权边。
#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;
const int maxn=1e5;
const int INF=1e9;
int dis[maxn],head[maxn],visit[maxn],outque[maxn];
int ip,que[maxn],flag;
struct node
{
int to;
int next;
int w;
};
node edge[maxn];
void add(int u,int v,int w)
{
edge[ip].to=v,edge[ip].w=w,edge[ip].next=head[u],head[u]=ip++;
}
void spfa(int s,int n)
{
memset(visit,0,sizeof(visit));
memset(outque,0,sizeof(outque));
for(int i=0;i<=n;i++)
dis[i]=INF;
dis[s]=0;
visit[s]=1;
int front=-1,tail=-1;
que[++tail]=s;
int top,to;
int temp;
while(front!=tail)
{
if(++front>n)
front-=n;
top=que[front];
outque[top]++;//入队次数加加
if(outque[top]>n)//入队次数大于n，存在负权环。算法结束
{
flag=0;
//printf("&&\n");
return;
}
visit[top]=0;
for(int k=head[top];k!=-1;k=edge[k].next)
{
to=edge[k].to;
temp=dis[top]+edge[k].w;
if(dis[to]>temp)
{
dis[to]=temp;
if(!visit[to])
{
if(++tail>n)
tail-=n;
que[tail]=to;
visit[to]=1;
}
}
}
}
}
int main()
{
int T,m,k,u,v,w,n;
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d",&n,&m,&k);
memset(head,-1,sizeof(head));
ip=0;
for(int i=0;i<m;i++)
{
scanf("%d%d%d",&u,&v,&w);
add(u,v,w);
add(v,u,w);
}
for(int i=0;i<k;i++)
{
scanf("%d%d%d",&u,&v,&w);
add(u,v,-w);
}
flag=1;
spfa(1,n);
if(!flag)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}