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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
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;
}
}
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原文地址：http://www.cnblogs.com/orchidzjl/p/4426351.html
