POJ3259 Wormholes SPFA 或者 bellman_ford

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 37539 Accepted: 13818

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

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.

求是否存在负权值回路

#include <iostream>#include <stdio.h>#include <string>#include <cstring>#include <algorithm>#include <cmath>#include <queue>const int INF = (1<<30);using namespace std;struct Node{    int v,w,nt;}f[100009];int cnt;int used[100009];int head[100009];int inq[100009];int dis[100009];queue<int>q;void add(int u,int v,int w){    f[cnt].v=u;    f[cnt].w=w;    f[cnt].nt=head[v];    head[v]=cnt++;}int spfa(int n){    for(int i=1;i<=n;i++)        dis[i]=INF;    dis[1]=0;    q.push(1);    used[1]++;    while(!q.empty())    {        int x=q.front(); q.pop();        inq[x]=0;        for(int u=head[x]; u!=-1; u=f[u].nt)        {            if(dis[f[u].v] > dis[x]+f[u].w)//如果可松弛则进行操作，否则不动            {                dis[f[u].v] = dis[x]+f[u].w;                if(!inq[f[u].v])               {                inq[f[u].v]=1;                if(++used[f[u].v]>n-1) return 0;                q.push(f[u].v);               }            }        }    }    return 1;}void init(int n){    cnt=0;    for(int i=1;i<=n;i++)    {        used[i]=0;        inq[i]=0;        head[i]=-1;    }}int main(){    int T;    int u,v,w;    int n,m,l;    scanf("%d",&T);    while(T--)    {        scanf("%d%d%d",&n,&m,&l);        init(n);        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<l;i++)        {            scanf("%d%d%d",&u,&v,&w);            add(u,v,-w);        }        if(!spfa(n)) puts("YES");        else puts("NO");    }    return 0;}

#include <iostream>#include <stdio.h>#include <string>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const int INF = (1<<26);struct Node{    int u,v,w;}f[10009];int cnt;int dis[10009];void add(int u,int v,int w){    f[cnt].u=u;    f[cnt].v=v;    f[cnt++].w=w;}int bellman_ford(int n){    for(int i=1;i<=n;i++) dis[i]=INF;    dis[1]=0;    int u,v,w;    for(int i=0;i<n-1;i++)        for(int j=0;j<cnt;j++)        {             u=f[j].u;             v=f[j].v;             w=f[j].w;            if(dis[u]<INF && dis[v]>dis[u]+w)            {                dis[v]=dis[u]+w;            }        }    for(int j=0;j<cnt;j++)        {             u=f[j].u;             v=f[j].v;             w=f[j].w;            if(dis[u]<INF && dis[v]>dis[u]+w)            {               return 0;            }        }        return 1;}int main(){    int T;    scanf("%d",&T);    int n,m,l;    while(T--)    {        scanf("%d%d%d",&n,&m,&l);        cnt=0;        int u,v,w;        for(int i=1;i<=m;i++)        {            scanf("%d%d%d",&u,&v,&w);            add(u,v,w);            add(v,u,w);        }        for(int i=0;i<l;i++)        {            scanf("%d%d%d",&u,&v,&w);            add(u,v,-w);        }        if(bellman_ford(n)==0) puts("YES");        else puts("NO");    }    return 0;}