Wormholes
Time Limit: 2000MSMemory Limit: 65536K
Total Submissions: 36717Accepted: 13438
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 toF (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 WLines 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 toE 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
70ms
#include<iostream> //79ms#include<cstdio>#include<cstring>#include<cmath>#define INF 10000000using namespace std;struct node{int u,v,w;} edge[5500];int low[5500];int n,m,z;int num=0;int Bellman(){for(int i=0; i<=n; i++)low[i]=INF;for(int i=0; i<n-1; i++){int flag=0;for(int j=0; j<num; j++){if(low[edge[j].u]+edge[j].w<low[edge[j].v]){low[edge[j].v]=low[edge[j].u]+edge[j].w;flag=1;}}if(flag==0) //存在负权回路break;}for(int j=0; j<num; j++) //判断负权回路{if(low[edge[j].u]+edge[j].w<low[edge[j].v])return 1;}return 0;}int main(){int T;scanf("%d",&T);while(T–){scanf("%d%d%d",&n,&m,&z);int a,b,c;num=0;for(int i=1; i<=m; i++){scanf("%d%d%d",&a,&b,&c);edge[num].u=a;edge[num].v=b;edge[num++].w=c;edge[num].u=b;edge[num].v=a;edge[num++].w=c;}for(int i=1; i<=z; i++){scanf("%d%d%d",&a,&b,&c);edge[num].u=a;edge[num].v=b;edge[num++].w=-c;}if(Bellman())printf("YES\n");elseprintf("NO\n");}}700ms
#include<iostream> //挨个点遍历#include<cstdio>#include<cstring>#include<cmath>#define INF 0x3f3f3f3fusing namespace std;struct node{int u,v,w;} edge[5500];int low[550];int n,m,z;int num=0;int Bellman(int u0){for(int i=0; i<=n; i++)low[i]=INF;low[u0]=0;for(int i=0; i<n; i++) //递推n次,让其构成环来判断{int flag=0;for(int j=0; j<num; j++){if(low[edge[j].u]!=INF&&low[edge[j].u]+edge[j].w<low[edge[j].v]){low[edge[j].v]=low[edge[j].u]+edge[j].w;flag=1;}}if(flag==0) //存在负权回路(减少时间)break;}if(low[u0]<0)return 1;return 0;}int main(){int T;scanf("%d",&T);while(T–){scanf("%d%d%d",&n,&m,&z);int a,b,c;num=0;for(int i=1; i<=m; i++){scanf("%d%d%d",&a,&b,&c);edge[num].u=a;edge[num].v=b;edge[num++].w=c;edge[num].u=b;edge[num].v=a;edge[num++].w=c;}for(int i=1; i<=z; i++){scanf("%d%d%d",&a,&b,&c);edge[num].u=a;edge[num].v=b;edge[num++].w=-c;}int biao=1;for(int i=1; i<=n; i++){if(Bellman(i)){printf("YES\n");biao=0;break;}}if(biao)printf("NO\n");}}/*780ms#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#define INF 0x3f3f3f3fusing namespace std;struct node{int u,v,w;} edge[5500];int low[5500];int n,m,z;int num=0;int Bellman(int u0){for(int i=0; i<=n; i++)low[i]=INF;low[u0]=0;//初始化for(int i=0; i<n-1; i++)//n-1次{int flag=0;for(int j=0; j<num; j++){if(low[edge[j].u]!=INF&&low[edge[j].u]+edge[j].w<low[edge[j].v]) //不同点{//存在low[edge[j].u]!=INF,就必须有low[u0]=0;初始化low[edge[j].v]=low[edge[j].u]+edge[j].w;flag=1;}}if(flag==0) //存在负权回路break;}for(int j=0; j<num; j++){if(low[edge[j].u]!=INF&&low[edge[j].u]+edge[j].w<low[edge[j].v])return 1;}return 0;}int main(){int T;scanf("%d",&T);while(T–){scanf("%d%d%d",&n,&m,&z);int a,b,c;num=0;for(int i=1; i<=m; i++){scanf("%d%d%d",&a,&b,&c);edge[num].u=a;edge[num].v=b;edge[num++].w=c;edge[num].u=b;edge[num].v=a;edge[num++].w=c;}for(int i=1; i<=z; i++){scanf("%d%d%d",&a,&b,&c);edge[num].u=a;edge[num].v=b;edge[num++].w=-c;}int biao=1;for(int i=1; i<=n; i++){if(Bellman(i)){printf("YES\n");biao=0;break;}}if(biao)printf("NO\n");}}*/
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生命不在长而在于好，只要每一次尽力的演示，都值得鼓励与喝采。
