USACO月赛2018年12月铜组_usaco 月赛 dfs-优快云博客

本文链接：https://blog.youkuaiyun.com/davidliule/article/details/115754568

这篇博客包含三段代码，分别涉及混合牛奶问题、桶列表管理和回溯算法。第一段代码解决了一个将三种牛奶倒入三个容器中使每个容器的牛奶量尽可能平均的问题。第二段代码处理了桶列表，找出在一系列操作后桶的最大累积值。第三段代码通过回溯算法求解6个数字的组合，使得在特定限制下数字总和变化的次数。

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T1 Mixing Milk

#include <iostream>
using namespace std;

int main() {
	int c[3], m[3];
	for (int i = 0; i < 3; i ++) cin >> c[i] >> m[i];
	
	for (int i = 0; i < 100; i ++) {
		int x = i % 3;
		int y = (i+1) % 3;
		
		if (m[x] >= c[y]-m[y]) m[x] -= c[y]-m[y], m[y] = c[y];
		else m[y] += m[x], m[x] = 0;
	}

	for (int i = 0; i < 3; i ++) cout << m[i] << endl;

	return 0;
}

T2 The Bucket List

#include <iostream>
using namespace std;

const int N = 1005;
int n, a[N], sum[N];

int main() {
	cin >> n;
	for (int i = 1; i <= n; i ++) {
		int s, t, b;
		cin >> s >> t >> b;
		
		a[s] += b;
		a[t+1] -= b;
	}

	int ans = 0;
	for (int i = 1; i <= 1000; i ++) {
		sum[i] = sum[i-1] + a[i];
		ans = max(ans, sum[i]);
	}	
	cout << ans << endl;

	return 0;
}

T3 Back and Forth

#include <iostream>
using namespace std;

int a[30], ttl = 1000;
bool b[30], mt[2010];

void dfs(int x) {
	if (x == 6) {
		mt[ttl] = true;
		return;
	}
	
	for (int i = 1; i <= 20; i ++) {
		if (x%2==0 && !b[i]) {
			ttl -= a[i];
			b[i] = true;
			dfs(x+1);
			b[i] = false;
			ttl += a[i];
		}
		else if (x%2==1 && b[i]) {
			ttl += a[i];
			b[i] = false;
			dfs(x+1);
			b[i] = true;
			ttl -= a[i];
		}
	}	
}

int main() {
	for (int i = 1; i <= 10; i ++) {
		cin >> a[i];
		b[i] = false;
	}
	for (int i = 11; i <= 20; i ++) {
		cin >> a[i];
		b[i] = true;
	}

	dfs(2);

	int cnt = 0;
	for (int i = 0; i <= 2000; i ++) {
		if (mt[i]) cnt ++;
	}
	cout << cnt;

	return 0;
}