CSU 1547: Rectangle

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.

题意：

    现有很多长方形，边长为a，b。其中a只有两种可能：1或2。    你需要找到一个足够大的长方形，边长为2，m。能把现有的长方形统统装进去。

分析：

    a为1的长方形，取尽可能接近的b的长方形一起放。    01背包问题解法，详见背包九讲。

AC代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;int T,n,ans,sum;int a[10005],b[10005];int dp[10005];int main(){    while(cin>>T)    {    while(T--)    {        cin>>n;        sum=0,ans = 0;        memset(a,0,sizeof(a));        memset(b,0,sizeof(b));        for(int i=0;i<n;i++)        {            cin>>a[i]>>b[i];            if(a[i]==1)            sum+=b[i];        }        for(int i=0;i<n;i++)        {            if(a[i]==2)            {                ans+=b[i];            }        }        int num = (sum>>1);        memset(dp,0,sizeof(dp));        for(int i=0;i<n;i++)        for(int j=num;j>=b[i];j--)        {            if(a[i]==1)            dp[j]=max(dp[j],dp[j-b[i]]+b[i]);        }        int aa=sum-dp[num];        aa=max(aa,dp[num]);        ans+=aa;        cout<<ans<<endl;    }}    return 0;}