POJ 3259 Wormholes

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 29315 Accepted: 10589

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.

跟上一个题一样 就是判断一个负环 难道最短路题都这样么 表示很费解 sad

#include <stdio.h>#include <string.h>#include <math.h>#include <stdlib.h>#include <ctype.h>#include <iostream>#include <algorithm>#include <map>#include <string>#include <vector>using namespace std;#define Cmp(a,b) strcmp(a,b)#define Copy(a,b) strcpy(a,b)#define Len(a) strlen(a)#define sn getchar()#define fr(a) for(int i=0;i<a;i++)#define fn(a) for(int i=1;i<=a;i++)#define f1(a) for(int i=1;i<a;i++)#define pn cout<<endl#define pc(a) printf("Case %d:",a)#define pr cout<<" "#define MEM(a) memset(a,0,sizeof(a))#define MEMF(a) memset(a,false,sizeof(a))#define w1 while(1)#define w(a) while(a--)#define INF 4294967295#define PI 3.14159265359const int dir4[4][2]= {{-1,0},{1,0},{0,-1},{0,1}};const int dir8[8][2]= {{-1,0},{1,0},{0,-1},{0,1},{-1,1},{1,-1},{-1,-1},{1,1}};/***********************************************/struct node{    int a,b,w;} Map[10001];int BF(int n,int s,int k){    int d[20002];    for(int i=1; i<=n; i++)        d[i]=INF;    d[s]=0;    for(int i=1; i<n; i++)    {        for(int j=0; j<k; j++)        {            int a=Map[j].a;            int b=Map[j].b;            int w=Map[j].w;            if(d[b]>d[a]+w)                d[b]=d[a]+w;        }    }    for(int j=0; j<k; j++)    {        int a=Map[j].a;        int b=Map[j].b;        int w=Map[j].w;        if(d[b]>d[a]+w)            return 1;    }    return 0;}int main(){    int n,m,w;    int a,b,c;    int T;    cin>>T;    while(T--)    {        MEM(Map);        cin>>n>>m>>w;        int k=0;        fr(m)        {            cin>>a>>b>>c;            Map[k].a=a;            Map[k].b=b;            Map[k++].w=c;            Map[k].a=b;            Map[k].b=a;            Map[k++].w=c;        }        fr(w)        {            cin>>a>>b>>c;            Map[k].a=a;            Map[k].b=b;            Map[k++].w=-c;        }        int ans=BF(n,1,k);        if(ans)            cout<<"YES"<<endl;        else            cout<<"NO"<<endl;    }    return 0;}