Rectangle 0-1背包

A - Rectangle

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%lld & %llu

Submit Status Practice CSU 1547

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

231 22 22 331 21 21 3

Sample Output

74

Hint

0-1背包问题。。背包大小 sum/2

//HDU1171：Big Event in HDU#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <algorithm>#include <iostream>using namespace std;int dp[100050];int f[105];int main() {int i,j,k,t,n,a,p,b;scanf("%d",&t);for (p=1;p<=t;p++){memset(dp,0,sizeof(dp));scanf("%d",&n);int ok=0;int sum=0;int len=0;for (j=1;j<=n;j++){scanf("%d%d",&a,&b);if (a==2){len+=b;}if (a==1){f[++ok]=b;sum+=b;}}for (i=1;i<=ok;i++){for (j=sum/2;j>=f[i];j--){dp[j]=max(dp[j],dp[j-f[i]]+f[i]);}}len+=sum-dp[sum/2];printf("%d\n",len);}return 0;}