题目

本题难度较低，操作比较简单，首先对于范围较小的N(<=100)，我们可以先跑一遍floyd，求出任意两点之间的最短路。对于很小的p(<=15),我们可以直接考虑全排列，运用到next_permutation(, )函数即可快速解决此题

Code：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
//Mystery_Sky
//
#define maxn 300
#define INF 0x3f3f3f3f
int point[maxn];
int road[maxn][maxn];
int vis[maxn];
int ans = INF, sum;
int n, w, a, p;
void dfs(int step, int dis, int get)
{   
    vis[step] = 1;
    if(get == p && step == n) {
        ans = min(dis, ans);
        return;
    }
    
    for(int i = 1; i <= n; i++) {
        if(i == step) continue;
        if(!vis[i] && point[i]) {
            dis += road[step][i];
            vis[i] = 1; 
            dfs(i, dis, get+1);
            vis[i] = 0;
            dis -= road[step][i];       
        }

        
    }
}

int main() {
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= n; j++) {
            scanf("%d", &w);
            road[i][j] = w;
        }
    }
    for(int k = 1; k <= n; k++) {
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= n; j++) {
                road[i][j] = min(road[i][j], road[i][k] + road[k][j]);
            }
        }
    }
    scanf("%d", &p);
    for(int i = 1; i <= p; i++) scanf("%d", &point[i]);
    sort(point+1, point+1+p);
    do {
        sum = road[1][point[1]];
        for(int i = 2; i <= p; i++) {
            sum += road[point[i-1]][point[i]];
        }
        sum += road[point[p]][n];
        ans = min(ans, sum);
    } while(next_permutation(point+1, point+p+1));//p!枚举全排列即可
    printf("%d\n", ans);
    return 0;
}

洛谷 P1690 贪婪的Copy

原文：https://www.cnblogs.com/Benjamin-cpp/p/10486041.html