一个人的旅行

Time Limit: 1000/1000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 31420????Accepted Submission(s): 10804

思路1：Dijkstra算法（链接附讲解）

AC代码：

#include<bits/stdc++.h>
#define N 1005
#define inf 9999999
using namespace std;

int e[N][N], dis[N],book[N];

void dijkstra(int n) {
    int i, j, k, Min;
    memset(book, 0, sizeof(book));
    //初始化某点的最短路径
    for (i = 0; i <= n; ++i)dis[i] = e[0][i];
    book[0] = 1;
    for (k = 1; k <= n; ++k) {
        Min = inf;
        for (i = 1; i <= n; ++i)
            if (!book[i] && dis[i] < Min)Min = dis[i], j = i;
        book[j] = 1;
        for (i = 1; i <= n; ++i)
            if (!book[i] && dis[i] > e[j][i] + dis[j])
                dis[i] = e[j][i] + dis[j];
    }
}
int main() {
    int t, s, d, Min, u, v, w, n;

    while (cin >> t >> s >> d) {
        memset(e, inf, sizeof(e));
        for (int i = 0; i < N; ++i)e[i][i] = 0;
        n = 0;
        //输入城市间路径与时间
        while (t--) {
            cin >> u >> v >> w;
            //因为有些路径是给了多次，取最小的
            if (e[u][v] > w)e[u][v] = e[v][u] = w;
            n = max(n, max(u, v));
        }
        //相邻城市距离为0
        while (s--) {
            cin >> v;
            e[0][v]= e[v][0] = 0;
        }
        dijkstra(n);
        Min = inf;
        while (d--) {
            cin >> v;
            Min = min(Min, dis[v]);
        }
        cout << Min << endl;
    }
    return 0;
}

#2066：一个人的旅行（Dijkstra算法入门题）

标签：mes   htm   最短路径   book   turn   limit   problem   杭电   pre   

原文地址：https://www.cnblogs.com/RioTian/p/12777514.html