poj 3259 Wormholes(Bellman-ford 算法)

题目地址

http://poj.org/problem?id=3259

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 37361 Accepted: 13736

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

这题主要是把题读懂。题意是：

2个test case 每个test case 第一行：  
N M W  
N个点 M条双向正权边 W条单向负权边    第一个test case 最后一行 3 1 3 是单向负权边，3->1的边权值是-3 

因为有负边，所以用bellman-ford算法就行。

AC代码：

#include <stdio.h>#include <string.h>#define N 505#define inf 100000int d[N];int n, m;typedef struct{    int from;    int to;    int cost;}node;node edges[5300];int bellman_ford(int m){    int i, j, k, t;    int ok;    for(i=1; i<=n; i++)        d[i] = inf;    d[1] = 0;    for(i=1; i<=n-1; i++)    {        ok = 1;        for(j=1; j<=m; j++)        {            if(d[edges[j].from] > d[edges[j].to] + edges[j].cost)            {                d[edges[j].from] = d[edges[j].to] + edges[j].cost;                ok = 0;            }        }        if(ok)            break;    }    for(j=1; j<=m; j++)    {        if(d[edges[j].from] > d[edges[j].to] + edges[j].cost)            return 0;    }    return 1;}int main(){    int i, j, t, x;    int a, b, c, w;    scanf("%d", &t);    while(t--)    {        scanf("%d%d%d", &n, &m, &w);        memset(edges, 0, sizeof(edges));        for(i=0, j=1; j<=m;j++)        {            scanf("%d%d%d", &a, &b, &c);            i++;            edges[i].from = a;            edges[i].to= b;            edges[i].cost = c;            i++;            edges[i].from = b;            edges[i].to = a;            edges[i].cost = c;        }        for(j=1; j<=w; j++)        {            scanf("%d%d%d", &a, &b, &c);            i++;            edges[i].from = a ;            edges[i].to = b;            edges[i].cost = -c;        }        if(bellman_ford(i))            printf("NO\n");        else            printf("YES\n");    }    return 0;}