POJ 3259 Wormholes (Bellman)

A - Wormholes

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status

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 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 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

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<map>#include<queue>using namespace std;const int INF=0x3f3f3f3f;const int MAXN=500+5;const int MAXM=5700+5;int u[MAXM];int v[MAXM];int w[MAXM];int d[MAXN];int MIN(int x,int y){return x>y?y:x;}int MAX(int x,int y){return x<y?y:x;}bool bellman(int n,int m,int t){int i,j,k;for(i=1;i<=n;i++)d[i]=INF;d[t]=0;for(i=1;i<n;i++)for(j=1;j<=m;j++)if(d[u[j]]+w[j]<d[v[j]])d[v[j]]=d[u[j]]+w[j];for(j=1;j<=m;j++)if(d[u[j]]+w[j]<d[v[j]])return 1;return 0;}int main(){//freopen("123.txt","r",stdin);int N,n,m,c;while(~scanf("%d",&N)){int e,i;while(N--){scanf("%d%d%d",&n,&m,&c);for(e=1;e<=m+c;e++){scanf("%d%d%d",&u[e],&v[e],&w[e]);if(e>m)w[e]*=-1;}for(e=m+c+1;e<=2*m+c;e++){u[e]=v[e-m-c];v[e]=u[e-m-c];w[e]=w[e-m-c];}if(bellman(n,2*m+c,1))printf("YES\n");else printf("NO\n");}}return 0;}

优化版本：

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<map>#include<queue>using namespace std;const int INF=0x3f3f3f3f;const int MAXN=500+5;const int MAXM=5700+5;int u[MAXM];int v[MAXM];int w[MAXM];int d[MAXN];int MIN(int x,int y){return x>y?y:x;}int MAX(int x,int y){return x<y?y:x;}bool bellman(int n,int m,int t){int i,j,k;for(i=1;i<=n;i++)d[i]=INF;d[t]=0;for(i=1;i<=n;i++){int flag=1;for(j=1;j<=m;j++)if(d[u[j]]+w[j]<d[v[j]]){d[v[j]]=d[u[j]]+w[j];flag=0;}if(flag)break;}if(i==n+1)return 1;return 0;}int main(){//freopen("123.txt","r",stdin);int N,n,m,c;while(~scanf("%d",&N)){int e,i;while(N--){scanf("%d%d%d",&n,&m,&c);for(e=1;e<=m+c;e++){scanf("%d%d%d",&u[e],&v[e],&w[e]);if(e>m)w[e]*=-1;}for(e=m+c+1;e<=2*m+c;e++){u[e]=v[e-m-c];v[e]=u[e-m-c];w[e]=w[e-m-c];}if(bellman(n,2*m+c,1))printf("YES\n");else printf("NO\n");}}return 0;}

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<map>#include<queue>using namespace std;const int INF=0x3f3f3f3f;const int MAXN=500+5;const int MAXM=5700+5;int u[MAXM];int v[MAXM];int w[MAXM];int first[MAXN];int next[MAXM];int d[MAXN];bool inq[MAXN];int num[MAXN];int MIN(int x,int y){return x>y?y:x;}int MAX(int x,int y){return x<y?y:x;}int cnt;void add_dedge(int u,int v,int w,int i){next[i]=first[u];first[u]=i;cnt++;}bool bellman(int n,int m,int t){int i;queue<int>q;memset(inq,0,sizeof inq);memset(num,0,sizeof num);for(i=1;i<=n;i++)d[i]=INF;d[t]=0;q.push(t);num[t]++;//inq[t]=1;while(!q.empty()){int e,x=q.front();q.pop();inq[x]=0;for(e=first[x];e!=-1;e=next[e]){if(d[x]+w[e]<d[v[e]]){d[v[e]]=d[x]+w[e];if(!inq[v[e]]){q.push(v[e]);num[v[e]]++;if(num[v[e]]>=n)return 1;inq[v[e]]=1;}}}}return 0;}int main(){//freopen("123.txt","r",stdin);int N,n,m,c;while(~scanf("%d",&N)){int e,i;while(N--){cnt=1;memset(first,-1,sizeof first);memset(next,-1,sizeof next);scanf("%d%d%d",&n,&m,&c);while(m--){scanf("%d%d%d",&u[cnt],&v[cnt],&w[cnt]);add_dedge(u[cnt],v[cnt],w[cnt],cnt);u[cnt]=v[cnt-1];v[cnt]=u[cnt-1];w[cnt]=w[cnt-1];add_dedge(u[cnt],v[cnt],w[cnt],cnt);}while(c--){scanf("%d%d%d",&u[cnt],&v[cnt],&w[cnt]);w[cnt]*=-1;add_dedge(u[cnt],v[cnt],w[cnt],cnt);}cnt--;if(bellman(n,cnt,1))printf("YES\n");else printf("NO\n");}}return 0;}