Rectangle(csu)

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

Hint

只能说经验不足，不知道这道题是0 1背包，背包大小 sum/2

记忆化搜索

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <iostream>
#include <set>
using namespace std;
int num[120];
int n;
int maxhave[10000][120];
int getmax(int sum,int n)
{
    int res;
    if(maxhave[sum][n] != -1) res = maxhave[sum][n];
    else if(n == 1){
        if(sum >= num[n]) res = num[n];
        else res = 0;
    }
    else if(sum >= num[n]){
        res = max(getmax(sum - num[n],n - 1) + num[n],getmax(sum,n - 1));
    }
    else res = getmax(sum,n - 1);
    maxhave[sum][n] = res;
    return res;
}
int main()
{
    int t;
    cin>>t;
    while(t--){
        memset(maxhave,-1,sizeof maxhave );
        cin>>n;
        int ans,sum;
        int a,b,c=1;
        ans = sum = 0;
        for(int i = 1; i <= n; ++i)
        {
            cin>>a>>b;
            if(a == 2) ans += b;
            else { num[c++] = b;sum += b; }
        }
        --c;
        int tmp = getmax(sum/2,c);
        ans = ans + max(tmp,sum-tmp);
        cout<<ans<<endl;
    }
}