POJ-3259 Wormholes(SPFA)

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 44207 Accepted: 16258

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.

Source

USACO 2006 December Gold

这道题一开始忘记贴代码了,有点尴尬...

题意:FJ在他的农场里发现有虫洞,每个虫洞可以让人从S传送到E,并且时间回退T,现在给出农场中有n个点,m条双向道路和w个单向虫洞连接这些点,询问能否从某个点出发,当他会到原点时,时间退回负值(注意:经过道路的时间是正值,经过虫洞的时间是负值)

题解:先用并查集处理下联通块,因为这不是单源最短路,所以要每个联通块都找一遍,然后用SPFA找负环,把时间设为距离,如果有负环,那么经过这个负环,时间肯定是回退的,设期望值为20,如果某个点进入队列超过这个期望值,就表示图里面有负环,FJ可以回到过去,输出YES,如果没有负环,则输出NO

#include<cstdio>#include<algorithm>#include<string.h>#include<queue>#include<vector>#include<functional>#include<iostream>using namespace std;const int maxn  = 505;const int maxn_edge = 6005;const int NN = 20;struct Edge{    int v,cost,nxt;}edge[maxn_edge];int head[maxn],tot,d[maxn],p[maxn],cnt[maxn];bool vis[maxn];void add(int u,int v,int cost){    edge[tot].v=v;    edge[tot].cost=cost;    edge[tot].nxt=head[u];    head[u]=tot++;   // cout<<u<<' '<<v<<' '<<cost<<' '<<endl;}void init(){    memset(d,0x3f,sizeof(d));    memset(vis,0,sizeof(vis));    memset(cnt,0,sizeof(cnt));}int find(int x){return p[x]==x?x:(p[x]=find(p[x]));}int main(){    int T,n,m,w;    freopen("in.txt","r",stdin);    scanf("%d",&T);    while(T--){        memset(head,-1,sizeof(head));        tot=0;        scanf("%d%d%d",&n,&m,&w);        for(int i=1;i<=n;i++) p[i]=i;        for(int i=0;i<m;i++){            int u,v,cost;            scanf("%d%d%d",&u,&v,&cost);            add(u,v,cost);            add(v,u,cost);            int p1=find(u),p2=find(v);            if(p1!=p2) p[p2]=p1;        }        for(int i=0;i<w;i++){            int u,v,cost;            scanf("%d%d%d",&u,&v,&cost);            add(u,v,-cost);            int p1=find(u),p2=find(v);            if(p1!=p2) p[p2]=p1;        }        bool flag=0;        for(int st=1;st<=n;st++){            if(flag) break;            if(find(st)!=st) continue;            init();            queue<int>q;            q.push(st);            d[st]=0;            while(!q.empty()){                if(flag) break;                int u=q.front();q.pop();                vis[u]=0;                for(int i=head[u];~i;i=edge[i].nxt){                    int v=edge[i].v;                    if(d[v]>d[u]+edge[i].cost){                        d[v]=d[u]+edge[i].cost;                        if(!vis[v]){                            vis[v]=1;                            cnt[v]++;                            q.push(v);                            if(cnt[v]>NN) {flag=1;break;}                        }                    }                }            }        }       // printf("%d\n",d[n]);        if(flag) printf("YES\n");        else printf("NO\n");    }    return 0;}