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最新推荐文章于 2022-10-09 23:28:27 发布

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最新推荐文章于 2022-10-09 23:28:27 发布

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本文链接：https://blog.youkuaiyun.com/katydid3018/article/details/47839795

本文深入探讨了信息技术领域的核心概念，包括前端开发、后端开发、移动开发、游戏开发、大数据开发、开发工具等多个方面，详细阐述了每个领域的关键技术和实际应用场景，旨在为读者提供全面的技术视野。

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Description

A set of n 1-dimensional items have to be packed in identical bins. All bins have exactly the same length l and each item i has length l_i $\le$ l . We look for a minimal number of bins q such that

each bin contains at most 2 items,
each item is packed in one of the q bins,
the sum of the lengths of the items packed in a bin does not exceed l .

You are requested, given the integer values n , l , l₁ , ..., l_n , to compute the optimal number of bins q .

Input

The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.

The first line of the input file contains the number of items n(1 $\le$ n $\le$ 10₅) . The second line contains one integer that corresponds to the bin length l $\le$ 10000 . We then have n lines containing one integer value that represents the length of the items.

Output

For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.

For each input file, your program has to write the minimal number of bins required to pack all items.

Sample Input

Sample Output

Note: The sample instance and an optimal solution is shown in the figure below. Items are numbered from 1 to 10 according to the input order.

$\epsfbox{p3503.eps}$

题目比较好看懂

每个箱子最多能装两个

最少要几个箱子

排序一下

最左最右相加

如果大于length

最后单独一个箱子

right--

如果不大于

两个一个箱子

right-- left++

#include <stdio.h>
#include <stdlib.h>
#define  N 100010

int item[N], length, n;
int cmp(const void *a, const void *b) {
	return (*(int *)a - *(int *)b);
}

int main(){
	int c;
	scanf("%d", &c);
	getchar();
	while (c--) {
		scanf("%d", &n);
		scanf("%d", &length);
		int count = 0;
		for (int i = 0; i < n; i++)
			scanf("%d", &item[i]);
		qsort(item, n, sizeof(item[0]), cmp);
		int left = 0, right = n - 1;
		while (left <= right) {
			if (left == right) {
				count++;
				break;
			}
			if ((item[left] + item[right]) > length) {
				right--;
			}
			else {
				left++, right--;
			}
			count++;
		}
		printf("%d\n", count);
		if (c)
			printf("\n");
	}
	return 0;
}