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.
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 backT seconds.
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 8Sample Output
NO YESHint
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.
t译文:农夫约翰在探索他的许多农场,发现了一些惊人的虫洞。虫洞是很奇特的,因为它是一个单向通道,可让你进入虫洞的前达到目的地!他的N(1≤N≤500)个农场被编号为1..N,之间有M(1≤M≤2500)条路径,W(1≤W≤200)个虫洞。作为一个狂热的时间旅行FJ的爱好者,他要做到以下几点:开始在一个区域,通过一些路径和虫洞旅行,他要回到最开时出发的那个区域出发前的时间。也许他就能遇到自己了:)。为了帮助FJ找出这是否是可以或不可以,他会为你提供F个农场的完整的映射到(1≤F≤5)。所有的路径所花时间都不大于10000秒,所有的虫洞都不大于万秒的时间回溯。
输入
第1行:一个整数F表示接下来会有F个农场说明。
每个农场第一行:分别是三个空格隔开的整数:N,M和W
第2行到M+1行:三个空格分开的数字(S,E,T)描述,分别为:需要T秒走过S和E之间的双向路径。两个区域可能由一个以上的路径来连接。
第M +2到M+ W+1行:三个空格分开的数字(S,E,T)描述虫洞,描述单向路径,S到E且回溯T秒。
输出
F行,每行代表一个农场
每个农场单独的一行,” YES”表示能满足要求,”NO”表示不能满足要求。
NO
YES
哈哈,要想回到原点。。。虫洞可以让 你回到过去,就相当于一个负权。。。然后判断是否存在负环,即走走走。。。然后又走回去了,,,
方法一:Bellmand---Ford
#include<stdio.h> #define INF 0x3f3f3f3f int dis[10000],bak[10000],u[10000],v[10000],t[10000],check,flag,n,m,w; int main() { int f; scanf("%d",&f); while(f--) { scanf("%d%d%d",&n,&m,&w); for(int i=1;i<=m;i++) scanf("%d%d%d",&u[i],&v[i],&t[i]); for(int i=m+1;i<=m+w;i++) { scanf("%d%d%d",&u[i],&v[i],&t[i]); //虫洞时间是负的·且单向 t[i]=-t[i]; } for(int i=1;i<=n;i++) dis[i]=INF; dis[1]=0; for(int k=1;k<n;k++) { for(int i=1;i<=n;i++)bak[i]=dis[i]; //将dis数组备份,可省时,提前跳出循环 for(int i=1;i<=m;i++) { if(dis[v[i]]>dis[u[i]]+t[i]) //农场两个方向都要判断 dis[v[i]]=dis[u[i]]+t[i]; if(dis[u[i]]>dis[v[i]]+t[i]) dis[u[i]]=dis[v[i]]+t[i]; } for(int i=m+1;i<=m+w;i++) { if(dis[v[i]]>dis[u[i]]+t[i]) //注意!!!!!!!!!!!!!从v[i]-u[i]是正的,方向别弄错了。。 dis[v[i]]=dis[u[i]]+t[i]; } check=0; for(int i=1;i<=n;i++) if(bak[i]!=dis[i]) { check=1; break; } if(check==0)break; } flag=0; for(int i=1;i<=m;i++) //虫洞不用判断,因为本身就是负的,判断m就好了.加上也行。 if(dis[v[i]]>dis[u[i]]+t[i])flag=1; if(flag) printf("YES\n"); else printf("NO\n"); } return 0; }方法二(floyd):
#include<cstdio> #include<cstring> #include<algorithm> #include<iostream> using namespace std; int map[505][505],n,m,k,num=0; int floyd() { int i,j,k,f=0; for(k=1;k<=n;k++) for(i=1;i<=n;i++){ for(j=1;j<=n;j++) { int t=map[i][k]+map[k][j]; if(map[i][j]>t)map[i][j]=t; /*map[i][j]=min(map[i][j],map[i][k]+map[k][j]);*/ //注意这里,用错了会超时!!!!不用min就跑了1688ms,,,, } if(map[i][i]<0)return 1; } return f; } int main() { int t; scanf("%d",&t); while(t--) { int i,j,a,b,c; scanf("%d%d%d",&n,&m,&k); memset(map,0x3f3f3f3f,sizeof(map)); for(i=1;i<=n;i++)map[i][i]=0; for(i=1;i<=m;i++) { scanf("%d%d%d",&a,&b,&c); if(c<map[a][b])map[a][b]=map[b][a]=c; } for(i=1;i<=k;i++) { scanf("%d%d%d",&a,&b,&c); map[a][b]=-c; } num++; int f=floyd(); if(!f)printf("NO\n"); else printf("YES\n"); } return 0; }