poj 3259 Wormholes（spfa） （spfa模板）

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 39978 Accepted: 14658

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, andW (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 toF (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 fromS toE that also moves the traveler backT 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.

Source

USACO 2006 December Gold

题目大意：

        给一个n个点的图，m条正权值边，n条负权值边，问是否存在一条回到从原点出发回到原点的路径使得权值为负。

解题思路：

        建边后跑spfa即可。

p.s.:

       1.注意正权值的是双向边，负权值的是单向边。

       2.注意判断负环是否存在，即统计每个点如队列次数是否>n

关于负权值的总结 

#include<stdio.h>#include<string.h>#include<queue>#define maxv 505#define maxe 6000//边数记得乘2#define C(a) memset(a,0,sizeof(a))#define C_1(a) memset(a,-1,sizeof(a))#define C_I(a) memset(a,0x3f,sizeof(a))#define INF 0x3f3f3f3fusing namespace std;struct node{int to,w,next;}e[maxe];int head[maxv],pos,d[maxv],cnt[maxv];int F,N,M,W,a,b,c;bool vis[maxv];bool spfa(int s){    C_I(d);C(cnt);C(vis);    d[s]=0;    queue<int>Q;    while(!Q.empty())Q.pop();    Q.push(s);    vis[s]=1;    while(!Q.empty())    {        int u=Q.front();Q.pop();vis[u]=0;        for(int i=head[u];i!=-1;i=e[i].next)        {            if(d[e[i].to]>d[u]+e[i].w)            {                d[e[i].to]=d[u]+e[i].w;                if(!vis[e[i].to])                {                    Q.push(e[i].to);                    vis[e[i].to]=1;                    cnt[e[i].to]++;                    if(cnt[e[i].to]>N)return 1;                }            }        }    }    return 0;}void add(int from,int to,int w){    e[pos]=(node){to,w,head[from]};    head[from]=pos++;}int main(){    scanf("%d",&F);    while(F--)    {        C(e),C_1(head);        pos=0;        scanf("%d%d%d",&N,&M,&W);        for(int i=0;i<M;i++)        {            scanf("%d%d%d",&a,&b,&c);            add(a,b,c);            add(b,a,c);        }        for(int i=0;i<W;i++)        {            scanf("%d%d%d",&a,&b,&c);            add(a,b,-c);        }        if(spfa(1)||d[1]<0)printf("YES\n");        else printf("NO\n");    }}