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 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 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.
题目大意：约翰的牧场有N个区域，这些区域直接有一些道路连接，另外约翰还发现一些虫洞，可以让他从S到E时间倒退T秒，问他能否走过若干路径之后回到原点看到出发前的自己？
思路：把带有虫洞的边看做负权边，若能看到出发前的自己即能找到负权环，用spfa判断一个节点重复n次以上出现在队列中即出现负权环。
code：
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define INF 0x3f3f3f3fint map[505][505];int first[1000010],dis[1000010];int next[1000010],point[1000010],w[1000010],q[1000010];int cnt[1000010];bool f[1000010];int n,m,ww,tot;void add(int x,int y,int c){ tot++; w[tot]=c; next[tot]=first[x]; first[x]=tot; point[tot]=y;}bool spfa(int s){ memset(dis,INF,sizeof(dis)); memset(f,0,sizeof(f)); memset(cnt,0,sizeof(cnt)); int l,r,x,y,k; l=100000; r=l; dis[s]=0; q[l]=s; f[s]=true; while (l<=r) { x=q[l]; if(cnt[x] >= n) return false; l++; k=first[x]; f[x]=false; while(k!=0) { y=point[k]; if(dis[y]>dis[x]+w[k]) { dis[y]=dis[x]+w[k]; if (!f[y]) { cnt[y]++; if (dis[y]<dis[q[l]]) { l--; q[l]=y; f[y]=true; } else { r++; q[r]=y; f[y]=true; } } } k=next[k]; } } return true;}int main(){ int t; scanf("%d",&t); while(t--) { tot = 0; memset(first,0,sizeof(first)); scanf("%d%d%d",&n,&m,&ww); int x,y,z; for(int i=1;i<=m;i++) { scanf("%d%d%d",&x,&y,&z); add(x,y,z); add(y,x,z); } for(int i=1;i<=ww;i++) { scanf("%d%d%d",&x,&y,&z); add(x,y,-z); add(x,y,-z); } if(spfa(1)) printf("NO/n"); else printf("YES/n"); } return 0;}新闻热点
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