Grass
题目大意:给你T条路的图,和S个起点和D个终点,问从S个起点中某个起点,到D个
终点中的某个终点的最短路径是多少。
思路:遍历起点S和终点D,用Dijkstra算法求单源最短路径即可。
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
const int MAXN = 1010;
int Map[MAXN][MAXN],Dist[MAXN];
bool vis[MAXN];
void Dijkstra(int N,int s)
{
    for(int i = 1; i <= N; ++i)
    {
        if(i != s)
            Dist[i] = Map[s][i];
    }
    Dist[s] = 0;
    vis[s] = true;
    for(int i = 1; i <= N-1; ++i)
    {
        int Min = 0xfffff0;
        int k = 0;
        for(int j = 1; j <= N; ++j)
        {
            if(!vis[j]  && Dist[j] < Min)
            {
                Min = Dist[j];
                k = j;
            }
        }
        if(k == 0)
            return;
        vis[k] = true;
        for(int j = 1; j <= N; ++j)
        {
            if(!vis[j] && Map[k][j] != 0xfffff0 && Dist[j] > Dist[k] + Map[k][j])
            {
                Dist[j] = Dist[k] + Map[k][j];
            }
        }
    }
}
int C1[MAXN],C2[MAXN];
int main()
{
    int T,S,D,N,a,b,w;
    while(~scanf("%d%d%d",&T,&S,&D))
    {
        for(int i = 1; i <= 1010; ++i)
            for(int j = 1; j <= 1010; ++j)
                Map[i][j] = 0xffffff0;
        memset(C1,0,sizeof(C1));
        memset(C2,0,sizeof(C2));
        N = 0;
        for(int i = 0; i < T; ++i)
        {
            scanf("%d%d%d",&a,&b,&w);
            N = max(N,a);
            N = max(N,b);
            if(Map[a][b] > w)
                Map[a][b] = Map[b][a] = w;
        }
        for(int i = 0; i < S; ++i)
        {
            scanf("%d",&C1[i]);
        }
        for(int i = 0; i < D; ++i)
        {
            scanf("%d",&C2[i]);
        }
        int Min = 0xffffff0;
        for(int i = 0; i < S; ++i)
        {
            memset(vis,false,sizeof(vis));
            memset(Dist,0,sizeof(Dist));
            Dijkstra(N,C1[i]);
            for(int j = 0; j < D; ++j)
                if(Dist[C2[j]] < Min)
                    Min = Dist[C2[j]];
        }
        printf("%d\n",Min);
    }
    return 0;
}
原文:http://blog.csdn.net/lianai911/article/details/42345905