short-path problem (Dijkstra) 分类： ACM TYPE 2014-09-01 23:51 111人阅读 评论(0) 收藏

#include <cstdio>
#include <iostream>
#include <cstring>

using namespace std;
const int INF = 0x3fffffff;

int  g[1005][1005];
int m;

int Dijkstra(int s,int t)
{
    bool visit[1005];
    int dis[1005];
    for(int i = 1; i <= m; ++i)
    {
        visit[i] = false;
        dis[i] = g[s][i];
    }
    visit[s] = true;

    int min = INF;
    int k;
    for(int i = 1; i < m; ++i)
    {
        min = INF;
        k = 1;
        for(int j = 1; j <= m; ++j)
        {
            if(!visit[j] && dis[j] < min)
            {
                min = dis[j];
                k = j;
            }
        }
        visit[k] = true;
        for(int j = 1; j <= m; ++j)
        {
            if(!visit[j] && dis[k] + g[k][j] < dis[j])
            {
                dis[j] = dis[k] + g[k][j];
            }
        }
    }
    return dis[t];
}

int main()
{
    int T, a, b, len;
    int s, t, p;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d", &m, &p);
        scanf("%d%d",&s,&t);
        for(int i = 1; i <= m; ++i)
            for(int j = 1; j < i; ++j)
                g[i][j] = g[j][i] = INF;

        while(p--)
        {
            scanf("%d%d%d", &a, &b, &len);
            if(g[a][b] > len)
                g[a][b] = g[b][a] = len;
        }

        printf("%d\n", Dijkstra(s,t));
    }
    return 0;
}

Code From Nyoj 115

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