2024 暑假友谊赛 1 （7.13）zhaosang

A-A

https://vjudge.net/contest/638765#problem/A
一开始贪心做不出来，后面发现是dp找到转移方程即可，01dp问题
代码如下

#include <bits/stdc++.h>

using namespace std;
using ll =long long;
ll v[10000010];
ll n;
ll ans;
ll prefix[10000010];
int main() {
	int N;
	cin >> N;
	vector<int> T(N);
	for (int i = 0; i < N; ++i) {
		cin >> T[i];
	}
	
	int all = accumulate(T.begin(), T.end(), 0);
	vector<bool> dp(all / 2 + 1, false);
	dp[0] = true;
	
	for (int i = 0; i < N; ++i) {
		for (int j = all / 2; j >= T[i]; --j) {
			if (dp[j - T[i]]) {
				dp[j] = true;
			}
		}
	}
	
	int result = all;
	for (int i = all / 2; i >= 0; --i) {
		if (dp[i]) {
			result = all - i;
			break;
		}
	}
	
	cout << result << endl;
	
	return 0;
}

B-B

板子题，

代码如下

#include <bits/stdc++.h>

using namespace std;
using ll =long long;
ll n;
ll ans;
ll prefix[10000010];
set<ll>st;

struct Point {
	ll x, y;
};

bool dfs(ll u, vector<vector<ll>> &adj, vector<bool> &visited, vector<ll> &match) {
	for (int v : adj[u]) {
		if (!visited[v]) {
			visited[v] = true;
			if (match[v] == -1 || dfs(match[v], adj, visited, match)) {
				match[v] = u;
				return true;
			}
		}
	}
	return false;
}

int maxMatching(vector<vector<ll>> &adj, int n) {
	vector<ll> match(n, -1);
	ll count = 0;
	for (int u = 0; u < n; ++u) {
		vector<bool> visited(n, false);
		if (dfs(u, adj, visited, match)) {
			count++;
		}
	}
	return count;
}

int main() {
	int N;
	cin >> N;
	
	vector<Point> red(N);
	vector<Point> blue(N);
	vector<vector<ll>> adj(N);
	for (int i = 0; i < N; ++i) {
		cin >> red[i].x >> red[i].y;
	}
	for (int i = 0; i < N; ++i) {
		cin >> blue[i].x >> blue[i].y;
	}	
	for (int i = 0; i < N; ++i) {
		for (int j = 0; j < N; ++j) {
			if (red[i].x < blue[j].x && red[i].y < blue[j].y) {
				adj[i].push_back(j);
			}
		}
	}
	
	ll result = maxMatching(adj, N);
	cout << result << endl;
	
	return 0;
}

H-H

https://vjudge.net/contest/638765#problem/H
签到题，不多说

#include <bits/stdc++.h>

using namespace std;
using ll =long long;
ll v[10000010];
ll n;
ll ans;
int main(){
	ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
	ll x,y,n;
	cin>>x>>y>>n;
	if(3*x<=y){
		cout<<n*x;
	}else {
		if(n>=3){
			cout<<(n/3)*y+(n%3)*x;
		}else
			cout<<n*x;
	}
}

其他题正在补。。。。。。
D-D

https://vjudge.net/contest/638765#problem/D

F-F

https://vjudge.net/contest/638765#problem/F