ZOJ 3905 Cake（dp）

Cake

Time Limit: 4 Seconds      Memory Limit: 65536 KB

Alice and Bob like eating cake very much. One day, Alice and Bob went to a bakery and bought many cakes.

Now we know that they have bought n cakes in the bakery. Both of them like delicious cakes, but they evaluate the cakes as different values. So they decided to divide those cakes by following method.

Alice and Bob do n / 2 steps, at each step, Alice choose 2 cakes, and Bob takes the cake that he evaluates it greater, and Alice take the rest cake.

Now Alice want to know the maximum sum of the value that she can get.

Input

The first line is an integer T which is the number of test cases.

For each test case, the first line is an integer n (1<=n<=800). Note that n is always an even integer.

In following n lines, each line contains two integers a[i] and b[i], where a[i] is the value of ith cake that Alice evaluates, and b[i] is the value of ith cake
that Bob evaluates. (1<=a[i], b[i]<=1000000)

Note that a[1], a[2]..., a
 are n distinct integers and b[1], b[2]..., b
 are n distinct integers.

Output

For each test case, you need to output the maximum sum of the value that Alice can get in a line.

Sample Input

1
6
1 6
7 10
6 11
12 18
15 5
2 14

Sample Output

28

Author: HUA, Yiwei

Submit    Status

/*
思路: 按 b排序，那么转化为了 在n个物品中取n/2个，同时这n/2个取完前面有n/2个物品被取了
我们算你一个人可以取的最小值，dp[i][j] 代表取到i这个位置，对方已经取了j 个物品
那么剩下来的物品应该满足 n-i>= (need-j)*2+1,

*/

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<set>
using namespace std;

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)
const int INF=0x3f3f3f3f;
typedef long long LL;

#define INF 0x3f3f3f3f
#define N 805

int dp

;
int n;
int va
;

struct stud{
int a,b;
bool operator<(const stud s)const
{
return b>s.b;
}
}f
;

void solve()
{
int i,j;
memset(dp,INF,sizeof(dp));
dp[0][0]=0;
int all=0;
for(i=1;i<=n;i++)
all+=va[i];

int need=n/2;

for(i=1;i<=need;i++)
for(j=i;j<=n;j++)
for(int k=i-1;k<j;k++)
{
if(n-j<=(need-i)*2) continue;
if(dp[k][i-1]>=INF) continue;
dp[j][i]=min(dp[j][i],dp[k][i-1]+va[j]);
}
int ans=INF;
for(int i=1;i<n;i++)
ans=min(ans,dp[i][need]);
printf("%d\n",all-ans);
}
int main()
{
int i,j,t;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(i=1;i<=n;i++)
scanf("%d%d",&f[i].a,&f[i].b);
sort(f+1,f+n+1);
for(i=1;i<=n;i++)
va[i]=f[i].a;
solve();
}
return 0;
}

/*

1 6 1 6 7 10 6 11 12 18 15 5 2 14
*/