pku Wormholes 第一周训练——最短路

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

第一周的题，已经是第二周末了。。还没做完。。唉。。效率啊。。。就是一个简单的判断有无负权值的环。

bellmen_ford做法

View Code

#include <iostream>
#include <cstdio>
#include <cstring>
#define maxn 5240
#define inf 0x7fffffff
using namespace std;

struct node
{
    int u,v;
    int w;
}g[maxn];
int n,m,w;
int dis[maxn];
bool bellman_ford(int len)
{
    int i,j;
    for (i = 0; i <= n; ++i)
    dis[i] = inf;
    int flag = 0;
    dis[1] = 0;
    for (i = 1; i < n; ++i)
    {
        for (j = 0; j < len; ++j)
        {
            if (dis[g[j].u] != inf && dis[g[j].u] + g[j].w < dis[g[j].v])//注意这里一定要判断dis[g[j].u] != inf
             dis[g[j].v] = dis[g[j].u] + g[j].w;
        }
    }
    for (i = 1; i < n; ++i)
    {
        if (flag) break;
        for (j = 0; j < len; ++j)
        {
             if (dis[g[j].u] + g[j].w < dis[g[j].v])
             {
                 flag = 1;
                 break;
             }
        }
    }
    if (flag) return true;
    else return false;
}
int main()
{
    int f,i,len;
    int x,y,wi;
    scanf("%d",&f);
    while (f--)
    {
        memset(g,0,sizeof(g));
        scanf("%d%d%d",&n,&m,&w);
        len = 0;
        for (i = 0; i < m; ++i)
        {
            scanf("%d%d%d",&x,&y,&wi);
            g[len].u = x;
            g[len].v = y;
            g[len].w = wi;
            len++;
            g[len].u = y;
            g[len].v = x;
            g[len].w = wi;
            len++;
        }
        for (i = 0; i < w; ++i)
        {
            scanf("%d%d%d",&x,&y,&wi);
            g[len].u = x;
            g[len].v = y;
            g[len].w = -wi;
            len++;
        }
        if (bellman_ford(len)) printf("YES\n");
        else               printf("NO\n");
    }
    return 0;
}