HDU 2066 一个人的旅行

http://acm.hdu.edu.cn/showproblem.php?pid=2066

题意：

输入数据有多组，每组的第一行是三个整数T，S和D，表示有T条路，和草儿家相邻的城市的有S个，草儿想去的地方有D个。 接着有T行，每行有三个整数a，b，time,表示a,b城市之间的车程是time小时；(1=<(a,b)<=1000;a,b 之间可能有多条路) 。

 接着的第T+1行有S个数，表示和草儿家相连的城市。

 接着的第T+2行有D个数，表示草儿想去地方。

思路：

需要注意一点，a，b可能有多条路，保存最短路。

#include<iostream> 
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;

#define INF 1000001

int T, S, D, n;

int d[1005][1005];
int e[1005];
int vis[1005], num[1005];


void Dijkstra()
{
    int min, pos;
    vis[0] = 1;
    for (int i = 0; i <= n; i++)
        num[i] = d[0][i];
    for (int i = 1; i <= n; i++)
    {
        min = INF;
        for (int j = 1; j <= n; j++)
        {
            if (num[j] < min && !vis[j])
            {
                pos = j;
                min = num[j];
            }
        }
        if (min == INF) break;
        vis[pos] = 1;
        for (int j = 1; j <= n; j++)
        {
            if (num[pos] + d[pos][j] < num[j] && !vis[j])
                num[j] = num[pos] + d[pos][j];
        }
    }
}

int main()
{
    //freopen("D:\txt.txt", "r", stdin);
    int a, b, t;
    while (~scanf("%d%d%d", &T, &S, &D))
    {
        n = 0;
        memset(vis, 0, sizeof(vis));
        for (int i = 0; i <= 1005; i++)
        {
            d[i][i] = 0;
            for (int j = i + 1; j <= 1005; j++)
                d[i][j] = d[j][i] = INF;
        }

        for (int i = 0; i < T; i++)
        {
            scanf("%d%d%d", &a, &b, &t);
            n = max(n, max(a, b));
            if (t<d[a][b])  d[a][b] = d[b][a] = t;  //这儿需要注意一下，因为两个点之间可能有多条路径，保存最短路
        }

        //把家和0点的距离设为0，这样从0出发肯定最先是家
        for (int i = 0; i < S; i++)
        {
            scanf("%d", &a);
            d[0][a] = d[a][0] = 0;
        }

        for (int i = 0; i < D; i++)
            scanf("%d", &e[i]);

        Dijkstra();

        int ans = INF;
        for (int i = 0; i < D; i++)
            ans = min(ans, num[e[i]]);
        printf("%d
", ans);
    }
    return 0;
}

本来是想用Floyd的，但是不行，超时了。

#include<iostream> 
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;

#define INF 1000001

int T, S, D;

int d[1005][1005];
int e[1005];
int vis[1005];
int p[1005];


int main()
{
    //freopen("D:\txt.txt", "r", stdin);
    int a, b, t;
    while (~scanf("%d%d%d", &T, &S, &D))
    {
        int cnt = 1;
        for (int i = 0; i <= 1000; i++)
        {
            d[i][i] = 0;
            for (int j = i + 1; j <= 1000; j++)
                d[i][j] = d[j][i] = INF;
        }
        p[0] = 0;
        for (int i = 0; i < T; i++)
        {
            scanf("%d%d%d", &a, &b, &t);
            if (!vis[a])
            {
                vis[a] = 1;
                p[cnt++] = a;
            }
            if (!vis[b])
            {
                vis[b] = 1;
                p[cnt++] = b;
            }
            if(t < d[a][b])  d[a][b] = d[b][a] = t;
        }

        for (int i = 0; i < S; i++)
        {
            scanf("%d", &a);
            d[0][a] = 0;
            d[a][0] = 0;
        }
        for (int i = 0; i < D; i++)
            scanf("%d", &e[i]);

        for (int k = 0; k < cnt; k++)
        for (int i = 0; i < cnt; i++)
        for (int j = 0; j < cnt; j++)
            d[p[i]][p[j]] = min(d[p[i]][p[j]], d[p[i]][p[k]] + d[p[k]][p[j]]);


        int ans = INF;
        for (int j = 0; j < D; j++)
            ans = min(ans, d[0][e[j]]);
        printf("%d
", ans);

        for (int i = 0; i < cnt; i++)
            vis[p[i]] = 0;
    }
    return 0;
}