Most Powerful （状态dp）

Recently, researchers on Mars have discovered N powerful atoms. All of them are different. These atoms have some properties. When two of these atoms collide, one of them disappears and a lot of power is produced. Researchers know the way every two atoms perform when collided and the power every two atoms can produce.

You are to write a program to make it most powerful, which means that the sum of power produced during all the collides is maximal.

Input

There are multiple cases. The first line of each case has an integer N (2 <= N <= 10), which means there are N atoms: A₁, A₂, ... , A_N. Then N lines follow. There are N integers in each line. The j-th integer on the i-th line is the power produced when A_i and A_j collide with A_j gone. All integers are positive and not larger than 10000.

The last case is followed by a 0 in one line.

There will be no more than 500 cases including no more than 50 large cases that N is 10.

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Output

Output the maximal power these N atoms can produce in a line for each case.

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Sample Input

2
0 4
1 0
3
0 20 1
12 0 1
1 10 0
0

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Sample Output

4
22

题目大概：

有n个原子，给出原子i和j碰撞后j原子迸发的能量，求最大的能量和。

思路：

把所有的原子压缩成一个数。

dp【i】状态为i时的最大能量和

方程  dp[i|tmp1]=max(dp[i|tmp1],dp[i]+mp[k][j])

代码注释

代码：

#include<cstdio>#include<iostream>#include<cstring>#define maxn (1<<10)using namespace std;int dp[maxn],mp[15][15];int main(){    int n,tot,ans;    while(scanf("%d",&n)!=EOF)    {        if(n==0) break;        for(int i=0;i<n;i++)        {            for(int j=0;j<n;j++)            {                scanf("%d",&mp[i][j]);            }        }        memset(dp,0,sizeof(dp));        tot=(1<<10);        for(int i=0;i<tot;i++)//枚举每个状态        {            for(int j=0;j<n;j++)//枚举爆发能量的原子            {                int tmp1=(1<<j);                if((i&tmp1)==0)//该原子要不包含在状态里                {                    for(int k=0;k<n;k++)//枚举和它碰撞的原子                    {                        int tmp2=(1<<k);                        if((i&tmp2)==0 && j!=k)//同样不包含                            dp[i|tmp1]=max(dp[i|tmp1],dp[i]+mp[k][j]);//转移方程，                    }                }            }        }        ans=0;        for(int i=0;i<tot;i++)        {            if(dp[i]>ans)                ans=dp[i];        }        printf("%d\n",ans);    }    return 0;}