初级->图算法->最短路径 poj 3259 Wormholes

// 题意 ： FJ有一些农场，这些农场里面有 N 块田地，田地里面有 W 个虫洞，田地和田地之间有路。

 

// 虫洞的功能： 时间倒流。

 

// 问FJ能不能看到他自己。有没有这样一条路径，所用时间为负值（当然要用到虫洞的性质时间才可能为负值）

// 思路： 首先建有向图，虫洞的时间设置为负值。

// 然后就判断这个图里面有没有负权回路就可以了 。 
  
// 因为负回路就可以满足条件，代表总共的需要的时间是负的，也就是时间倒流了，那么FJ的目标就实现了呗。

 

 

 

#include<stdio.h>#include<string.h>#define MAX 9999999int n,m,w;int dist[501];struct node{    int x,y,w;}edg[5401];int bellman_ford(){    int i,j,flag;    m=2*m+w;    for(i=1;i<n;i++)    {        flag=0;        for(j=1;j<=m;j++)        if(dist[edg[j].y]>dist[edg[j].x]+edg[j].w)        {            dist[edg[j].y]=dist[edg[j].x]+edg[j].w;            flag=1;        }        if(!flag) break;    }    for(j=1;j<=m;j++)    if(dist[edg[j].y]>dist[edg[j].x]+edg[j].w)    return 1;    return 0;}int main(){    int i,j,a,b,c,F;    scanf("%d",&F);    while(F--)    {        scanf("%d%d%d",&n,&m,&w);        for(j=1,i=1;i<=m;i++)        {            scanf("%d%d%d",&a,&b,&c);            edg[j].x=a;            edg[j].y=b;            edg[j].w=c;            j++;            edg[j].x=b;            edg[j].y=a;            edg[j].w=c;            j++;        }        for(i=1;i<=w;i++)        {            scanf("%d%d%d",&a,&b,&c);            edg[j].x=a;            edg[j].y=b;            edg[j].w=-c;            j++;        }        if(bellman_ford())        printf("YES\n");        else        printf("NO\n");    }    return 0;}

 

 

 

 

Wormholes

Time Limit: 2000MSMemory Limit: 65536KTotal Submissions: 21843Accepted: 7784

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 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 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