「CF852D」Exploration Plan

题目描述

给定一张 \(V\) 个点，\(M\) 条边的边带权无向图，有 \(N\) 个人分布在图上的点上，第 \(i\) 个人在 \(x_i\) 这个点上，定义从一个点走到另一个点的时间为所走的路径上所有边权之和，问至少过多久才可以满足至少有 \(K\) 个点上有人。

数据范围：

\(1\le V \le600,1\le E \le 20000,1\le N \le\min(V,200),1\le K \le N\)

基本思路

首先可以二分答案。

对于当前二分到的 \(mid\)，我们对于每一个人，都向他可以去到的点连一条边（路径的最短距离可以用 \(\text{Floyd}\) 预处理一下)，然后直接跑二分图最大匹配就好了。

细节注意事项

• 记得判无解

参考代码

/*--------------------------------
  Author: The Ace Bee
  Blog: www.cnblogs.com/zsbzsb
  This code is made by The Ace Bee
--------------------------------*/
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cctype>
#include <cmath>
#include <ctime>
#include <queue>
#define rg register
using namespace std;
template < typename T > inline void read(T& s) {
	s = 0; int f = 0; char c = getchar();
	while (!isdigit(c)) f |= (c == '-'), c = getchar();
	while (isdigit(c)) s = s * 10 + (c ^ 48), c = getchar();
	s = f ? -s : s;
}

const int _ = 700;
const int INF = 2147483647;

int n, m, p, k, x[_];
int dis[_][_], vis[_], bel[_], g[_][_];

inline int dfs(int u) {
	for (rg int i = 1; i <= n; ++i) {
		if (vis[i] || !g[u][i]) continue;
		vis[i] = 1;
		if (bel[i] == 0 || dfs(bel[i]))
			return bel[i] = u, 1;
	}
	return 0;
}

inline bool check(int mid) {
	memset(g, 0, sizeof g);
	for (rg int i = 1; i <= p; ++i)
		for (rg int j = 1; j <= n; ++j)
			g[i][j] = (int) dis[x[i]][j] <= mid;
	int res = 0;
	memset(bel, 0, sizeof bel);
	for (rg int i = 1; i <= p; ++i)
		memset(vis, 0, sizeof vis), res += dfs(i);
	return res >= k;
}

int main() {
#ifndef ONLINE_JUDGE
	freopen("in.in", "r", stdin);
#endif
	read(n), read(m), read(p), read(k);
	for (rg int i = 1; i <= p; ++i) read(x[i]);
	for (rg int i = 1; i <= n; ++i)
		for (rg int j = 1; j <= n; ++j)
			dis[i][j] = 1e9;
	for (rg int i = 1; i <= n; ++i) dis[i][i] = 0;
	for (rg int u, v, d, i = 1; i <= m; ++i) {
		read(u), read(v), read(d);
		dis[v][u] = dis[u][v] = min(dis[u][v], d);
	}
	for (rg int k = 1; k <= n; ++k)
		for (rg int i = 1; i <= n; ++i)
			for (rg int j = 1; j <= n; ++j)
				dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]);
	int l = 0, r = 1731311 + 1;
	while (l < r) {
		int mid = (l + r) >> 1;
		if (check(mid)) r = mid;
		else l = mid + 1;
	}
	if (l > 1731311) puts("-1");
	else printf("%d\n", l);
	return 0;
}

完结撒花 \(qwq\)