Poj 3259 wormholes (spfa)

Description

While processing his own 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 comprisesN(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 mapsF(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. FFarm 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 SAnd EThat requires TSeconds 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 STo EThat also moves the traveler back TSeconds.

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 cocould travel back in time by the cycle 1-> 2-> 3-> 1, arriving back at his starting location 1 second before he leaves. he cocould start from anywhere on the cycle to accomplish this.

Practice spfa.

Question: N places, M roads, and W wormhole holes. Determine whether the path can be time-backed.

Spfa: determines whether the number of teams has exceeded n. Bellman_ford directly determines.

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<limits.h>#include<queue>using namespace std;const int INF=0x3f3f3f3f;const int maxn=3000*2;int end[maxn],cost[maxn];int next[maxn],cnt[maxn],head[maxn];int t,n,m,w,e;bool flag;int d[maxn],visit[maxn];void add(int u,int v,int w){    end[e]=v;    cost[e]=w;    next[e]=head[u];    head[u]=e++;}void SPFA(){    queue<int>q;    memset(visit,0,sizeof(visit));    memset(cnt,0,sizeof(cnt));    memset(d,INF,sizeof(d));    visit[1]=1;    cnt[1]++;    d[1]=0;    q.push(1);    while(!q.empty())    {        int uu=q.front();        q.pop();        visit[uu]=0;        for(int i=head[uu];i!=-1;i=next[i])        {            int vv=end[i];            int ww=cost[i];            if(d[vv]>d[uu]+ww)            {                d[vv]=d[uu]+ww;                if(!visit[vv])                {                    visit[vv]=1;                    q.push(vv);                    if(++cnt[vv]>=n)                    {                        flag=false;                        return ;                    }                }            }        }    }    return ;}int main(){   int x,y,z;   scanf("%d",&t);   while(t--)   {       scanf("%d%d%d",&n,&m,&w);       e=0;       memset(head,-1,sizeof(head));       for(int i=0;i<m;i++)       {           scanf("%d%d%d",&x,&y,&z);           add(x,y,z);           add(y,x,z);       }       for(int i=0;i<w;i++)       {           scanf("%d%d%d",&x,&y,&z);           add(x,y,-z);       }       flag=true;       SPFA();       if(!flag)   printf("YES\n");       else   printf("NO\n");   }   return 0;}

Bellman_ford:

#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>using namespace std;#define MAXN 3000*2#define INF 0xFFFFFFFint t , n , m, w;int dis[MAXN];struct Edge{   int x;   int y;   int value;}e[MAXN];bool judge(){     for(int i = 0 ; i < m*2+w ; i++){        if(dis[e[i].y] > dis[e[i].x] + e[i].value)          return false;     }     return true;}void Bellman_Ford(){     dis[1] = 0;     for(int i = 2 ; i <= n ; i++)        dis[i] = INF;     for(int i = 1 ; i <= n ; i++){        for(int j = 0 ; j < m*2+w; j++){           if(dis[e[j].y] > dis[e[j].x] + e[j].value)             dis[e[j].y] = dis[e[j].x] + e[j].value;        }     }     if(judge())       printf("NO\n");     else       printf("YES\n");}int main(){   int uu,vv,ww,i;   scanf("%d",&t);   while(t--)   {       scanf("%d%d%d",&n,&m,&w);       for(i=0;i<m*2;)       {           scanf("%d%d%d",&uu,&vv,&ww);           e[i].x=uu;           e[i].y=vv;           e[i++].value=ww;           e[i].x=vv;           e[i].y=uu;           e[i++].value=ww;       }       for(;i<m*2+w;i++)       {           scanf("%d%d%d",&uu,&vv,&ww);           e[i].x=uu;           e[i].y=vv;           e[i].value=-ww;       }       Bellman_Ford();   }   return 0;}