USACO Bronze 2024 February Problem 2 Milk Exchange

2024-02-20 23:10:02
开发
50

Problem

Farmer John's $N$ ( $1 \leq N \leq 2*10^5$ ) cows are lined up in a circle such that for each $i$ in $1, 2, ..., N - 1$ , the cow to the right of cow $i$ is cow $i + 1$ , and the cow to the right of cow $N$ is cow $1$ . The $i$ th cow has a bucket with integer capacity $a_i$ ( $1 \leq a_i \leq 10^9$ ) liters. All buckets are initially full.

Every minute, the cows exchange milk according to a string $s_1s_2...s_N$ consisting solely of the characters ‘ $L$ ’ and ‘ $R$ ’. if the $i$ th cow has at least $1$ liter of milk, she will pass $1$ liter of milk to the cow to her left if $s_i = `L$ ’ , or to the right if $s_i = `R$ ’. All exchanges happen simultaneously (i.e., if a cow has a full bucket but gives away a liter of milk but also receives a liter, her milk is preserved). If a cow's total milk ever ends up exceeding $a_i$ , then the excess milk will be lost.

FJ wants to know: after $M$ minutes ( $1 \leq M \leq 10 ^ 9$ ), what is the total amount of milk left among all cows?

INPUT FORMAT (input arrives from the terminal / stdin):

The first line contains $N$ and $M$ .

The second line contains a string $s_1s_2...s_N$ consisting solely of the characters ‘ $L$ ’ or ‘ $R$ ’, denoting the direction each cow will pass their milk in.

The third line contains integers $a_1,a_2,...,a_N$ , the capacities of each bucket.

OUTPUT FORMAT (print output to the terminal / stdout):

Output an integer, the sum of milk among all cows after $M$ minutes.

Note that the large size of integers involved in this problem may require the use of 64-bit integer data types (e.g., a "long long" in C/C++).

SAMPLE INPUT:

3 1
RRL
1 1 1

SAMPLE OUTPUT:

Cows $2$ and $3$ pass each other one liter of milk, so their milk is preserved. When cow $1$ passes their milk to cow $2$ , cow $2$ 's bucket overflows, and one liter of milk is lost after one minute.

SAMPLE INPUT:

5 20
LLLLL
3 3 2 3 3

SAMPLE OUTPUT:

Each cow is passing a liter of milk to the cow on the left and gaining a liter of milk from the cow on the right, so all of the milk is preserved regardless of how much time passes.

SAMPLE INPUT:

9 5
RRRLRRLLR
5 8 4 9 3 4 9 5 4

SAMPLE OUTPUT:

Initially, there are a total of 51 liters of milk. After 5 minutes, cows $3$ , $6$ , and $7$ will lose 5, 3, and 5 liters of milk respectively. Therefore, a total of 38 liters of milk remain.

SCORING:

Inputs 4-8: $N,M \leq 1000$
Inputs 9-16: No additional constraints.

Solution

算法与解法

（讲得有可能不清楚，请谅解）

观察题目，发现 $R$ 倒入右边桶内， $L$ 倒入左边桶内，那如果当序列中出现了 $RL$ ，就代表左边奶牛会给右边 $1L$ ，右边奶牛会给左边 $1L$ ，就代表这两个奶牛的瓶子一直保持着满瓶的状态，如果一旦外面有牛奶倒入，就会使瓶子满杯而产生总量损失，也只有这种情况会产生损失。外面有任何牛奶倒入就代表：左边临近的奶牛字符为 $R$ ，后边临近的奶牛字符为 $L$ 。这就说明左/右一定会浪费牛奶，浪费牛奶数应为 $min(m, a_i)$ ，因为倒牛奶最多有 $m$ 轮，每轮只能给予一升牛奶， $mL$ 便是最大值了，但如果奶牛已没有奶了，便无法给予。

但对于每个 $RL$ 输入的数值不仅取决于临近的 $R$ 与 $L$ ，也可能有一下情况：

RRRRRLLLLL

我们会发现中间有一个 $RL$ ，而左边是 4 个 $R$ ，这会引发一个奇妙的现象：最左边的 $R$ 每轮给予下一个 $R$ 一升，下一个 $R$ 会给予再下一个 $R$ 一升，引发了一个接力，直到将牛奶给予了 $RL$ 。出现了这个“接力”现象，计算浪费牛奶数就有亿点儿困难了，我们可以发现最后一个奶牛会经过接力持续给予牛奶，当没有牛奶时，就需要倒数第二个奶牛给予牛奶；当倒数第二个奶牛没有牛奶时，就需要倒数第三个奶牛给予牛奶……以此类推。根据这个现象，我们可以找到解决方法：从 $RL$ 处往左开始遍历，要求 $R$ 不断开，则每一个就是 $sum = min(sum + a_i, m)$ 。其实就是将附近的 $R$ 奶牛所持有的牛奶数累加，最大值不超过 $m$ ，找出一共可以给予 $RL$ 多少升牛奶（也就是损失了多少牛奶）。 $L$ 与以上做法同理，只不过换了一个遍历方向。

综上所述，我们需要找到每个 $RL$ ，被找到给 $RL$ 提供牛奶的奶牛，计算损失数，把所有损失数从总牛奶数减去，即为答案。

特别要提醒的是：第一只奶牛与第 $n$ 只奶牛是连起来的，在遍历和寻找 $RL$ 请注意这一点。

代码

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N = 2e5 + 1;
int n, m, res, a[N];
char c[N];
signed main(){
	scanf("%lld%lld%s", &n, &m, c);
	for(int i = 1;i <= n; i++){
		scanf("%lld", &a[i]);
		res += a[i];						//计算总牛奶值 
	}
	if(c[0] == 'L' && c[n - 1] == 'R')		//第一个与第 n 个形成 RL
		c[0] = ' ', c[n - 1] = ' ';
	for(int i = 1;i < n; i++)
		if(c[i - 1] == 'R' && c[i] == 'L')	//i-1 与 i 形成 RL
			c[i - 1] = ' ', c[i] = ' ';		//用空格代替，以记录 
	for(int i = 0;i < n; i++){
		int suma = 0, sumb = 0;				//suma 记录左边相连的奶牛给予牛奶的数量，sumb 则是记录右边的 
		int l = i - 1;
		if(l < 0)
			l = n - 1;
		if(c[l] != ' ' || c[i] != ' ')
			continue ;
		bool flag = false;
		for(int j = l - 1;j >= 0 && !flag; j--){
			if(c[j] == 'R')					//是否相连 
				suma = min(suma + a[j + 1], m);
			else
				flag = true;
		}
		if(!flag){							//如果持续相连直到第一个，那就从第 n 个往前检查是否还有相连 
			for(int j = n - 1;j >= 0 && !flag; j--){
				if(c[j] == 'R')
					suma = min(suma + a[j + 1], m);
				else
					flag = true;
			}
		}
		flag = false;						//以下同 suma 
		for(int j = i + 1;j < n && !flag; j++){
			if(c[j] == 'L')
				sumb = min(sumb + a[j + 1], m);
			else
				flag = true;
		}
		if(!flag){
			for(int j = 0;j < n && !flag; j++){
				if(c[j] == 'L')
					sumb = min(sumb + a[j + 1], m);
				else
					flag = true;
			}
		}
		res -= suma + sumb;					//从总牛奶数减去此 RL 损失数
	}
	printf("%lld", res);
	return 0;
}

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