郊区春游

有详细注释

#include <bits/stdc++.h>

#define INFI 9999999
#define FIO ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
using namespace std;
typedef long long int ll;
int n, m, place_num, road[209][209], want_go[20];
ll dp[1 << 15][20], ans;

int main() {
    FIO;
    cin >> n >> m >> place_num;
    //place_num总共想要去的地点数
    for (int i = 1; i <= place_num; i++) {
        cin >> want_go[i];
    }
    //初始化不同点之间的距离为无穷大
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            if (i != j) {
                road[i][j] = INFI;
            }
        }
    }
    int a, b, c;
    for (int i = 1; i <= m; i++) {
        cin >> a >> b >> c;
        road[a][b] = min(road[a][b], c);
        road[b][a] = road[a][b];
    }
    //Floyd求最短路
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            for (int k = 1; k <= n; k++) {
                road[j][k] = min(road[j][k], road[j][i] + road[i][k]);
            }
        }
    }
    //初始化dp数组
    memset(dp, INFI, sizeof(dp));
    for (int i = 1; i <= place_num; i++) {
        dp[1 << (i - 1)][i] = 0;
    }
    for (int i = 1; i < (1 << place_num) - 1; i++) {
        //转换成二进制，1表示已经去了该地点，0表示没有去过该地点
        for (int j = 1; j <= place_num; j++) {
            if (i & (1 << (j - 1))) {//i的二进制的第j位为1，即已经去了第j个顶点
                //那么我们就可用当前的数据去更新还未去的地点的数据
                for (int k = 1; k <= place_num; k++) {
                    if ((i & (1 << (k - 1))) == 0) {//i的二进制的第k位为0
                        dp[i + (1 << (k - 1))][k] = min(dp[i + (1 << (k - 1))][k],
                                                        dp[i][j] + road[want_go[j]][want_go[k]]);
                        //dp数组第一维存储i，记录了当前已经去了和没去的地点
                        //第二维存储当前走法的最后到达的位置
                    }
                }
            }
        }
    }
    ans = INFI;
    for (int i = 1; i <= place_num; i++) {
        ans = min(ans, dp[(1 << place_num) - 1][i]);
    }
    cout << ans << endl;
    return 0;
}