CSU--1547: Rectangle【01背包】

1547: Rectangle

      Time Limit: 1 Sec     Memory Limit: 256 Mb     Submitted: 1281     Solved: 366    

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Description

Now ,there are some rectangles. The area of these rectangles is 1* x or 2 * x ,and now you need find a big enough rectangle( 2 * m) so that you can put all rectangles into it(these rectangles can't rotate). please calculate the minimum m satisfy the condition.

Input

There are some tests ,the first line give you the test number.
Each test will give you a number n (1<=n<=100)show the rectangles number .The following n rows , each row will give you tow number a and b. (a = 1 or 2 , 1<=b<=100).

Output

Each test you will output the minimum number m to fill all these rectangles.

Sample Input

2
3
1 2
2 2
2 3
3
1 2
1 2
1 3

Sample Output

7
4

解题思路：这个题目讲的是会有1*，2*这2种长方形，将这些长方形都放在一个2*的长方形，让我们找出最小的长方形，2*的我们就直接将长度加起来，主要的是如何处理1*的长方形，我们自己去画图的时候我们会发现我们把1*的尽量是1*的总和的一半，尽量是一半，这样肯定是最短的，具体处理就是用01背包处理，背包的容量是sum/2，vo和va都是b，然后找最大的就好，在最后输出答案的时候还要ans[sum/2]和sum-ans[sum/2]中选最大的，具体的自己画图理解理解

#include<bits/stdc++.h>
using namespace std;
int va[102],vo[102],ans1[10002];//ans1这个数组一开始开小了，出现了段错误
int main(void)
{
	int t,n,a,b,sum;
	scanf("%d",&t);
	while(t--)
	{
		int k=0,ans=0,sum=0;
		scanf("%d",&n);
		memset(ans1,0,sizeof ans1);
 		for(int i=0;i<n;i++)
		{
			scanf("%d%d",&a,&b);
			if(a==2) ans+=b;
			else
			{
				sum+=b;
				va[k]=b;
				vo[k]=b;
				k++;
			}
		}
		int v=sum/2;
		for(int i=0;i<k;i++)
		{
			for(int j=v;j>=vo[i];j--)
			{
				ans1[j]=max(ans1[j],ans1[j-vo[i]]+va[i]);
			}
		}
		printf("%d\n",ans+max(ans1[v],sum-ans1[v]));	
	}
	return 0;
}