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.
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).
Each test you will output the minimum number m to fill all these rectangles.
2
3
1 2
2 2
2 3
3
1 2
1 2
1 3
7
4
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
#define Mod 1000000007
#define ll long long
#define N 111
#define INF 1010010010
using namespace std;
struct node {
int x;
int y;
} a[N];
int n;
int cmp(node a,node b) {
if(a.x==b.x)return a.y>b.y;
return a.x>b.x;
}
int dp[N*N];
int main() {
//freopen("in.txt","r",stdin);
int t;
cin>>t;
while(t--) {
scanf("%d",&n);
for(int ii=0; ii<n; ii++) {
scanf("%d%d",&a[ii].x,&a[ii].y);
}
sort(a,a+n,cmp);
a[n].y=0;
int ans=0;
int i=0;
while(i<n&&a[i].x==2) {
ans+=a[i].y;
i++;
}
int sum=0;
for(int ii=i; ii<n; ii++)sum+=a[ii].y;
int s=sum/2;
memset(dp,0,sizeof dp);
for(int ii=i;ii<n;ii++)
{
for(int j=s;j>=a[ii].y;j--)
{
dp[j]=max(dp[j],dp[j-a[ii].y]+a[ii].y);
}
}
int ss=sum-dp[s];
printf("%d\n",ans+max(ss,dp[s]));
}
return 0;
}原文地址：http://blog.csdn.net/acm_baihuzi/article/details/45798941
