本文主要是介绍neuq-acm预备队训练week 8 P8794 [蓝桥杯 2022 国 A] 环境治理，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
题目描述
输入格式
输出格式
输出一行包含一个整数表示答案。
输入输出样例
解题思路
最短路+二分
AC代码
#include<bits/stdc++.h>
using namespace std;
long long temp,n, Q;
long long f[105][105],min_f[105][105],cut[105],dis[105][105];//cut为减少多少，dis表示减少后的
bool check(long long day)
{temp = 0;//赋初值long long times = day / n;long long now = day%n;for (int i = 1; i <= n; i++){if (i <= now)cut[i] = times + 1;else cut[i] = times;}for (int i = 1; i <= n; i++){for (int j = 1; j <= n; j++){dis[i][j] = max(f[i][j] - cut[i] - cut[j], min_f[i][j]);}}for (int k = 1; k <= n; k++)//Floyd模板for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++)dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]);for (int i = 1; i <= n; i++){for (int j = 1; j <= n; j++){temp += dis[i][j];}}if (temp <= Q)return true;else return false;
}
int main()
{cin >> n>>Q;for (int i = 1; i <= n; i++){for (int j = 1; j <= n; j++){cin >> f[i][j];}}for (int i = 1; i <= n; i++){for (int j = 1; j <= n; j++){cin >> min_f[i][j];}}long long left = 0, right = 1e9,mid;while (left<right){mid = (left + right) / 2;if (check(mid)) right = mid;else left = mid + 1;}if (left == 1e9)cout << "-1" << endl;else cout << left << endl;return 0;
}这篇关于neuq-acm预备队训练week 8 P8794 [蓝桥杯 2022 国 A] 环境治理的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
