ZOJ 4257 Most Powerful （状态压缩DP）

Most Powerful

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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_jgone. 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.

Output

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

Sample Input

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

Sample Output

4
22

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Author: GAO, Yuan
Contest: ZOJ Monthly, February 2011

简单的状态压缩DP，

状态用二进制表示，0表示没有消除，1表示已经消除了。

dp[i]表示状态为i时获得的最大值。然后不断枚举在的两个进行碰撞。

（一开始想了个错误的思路，555555555）

//============================================================================
// Name        : ZOJ.cpp
// Author      : 
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int dp[1<<12];
int g[12][12];
int main()
{
//    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
    int n;
    while(scanf("%d",&n)==1 && n)
    {
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
                scanf("%d",&g[i][j]);
        memset(dp,-1,sizeof(dp));
        dp[0]=0;
        int tot=(1<<n);
        int ans=0;
        for(int i=0;i<tot;i++)
        {
            if(dp[i]==-1)continue;
            ans=max(ans,dp[i]);
            for(int j=0;j<n;j++)
                for(int k=0;k<n;k++)
                    if(j!=k && (i&(1<<j))==0 && (i&(1<<k))==0)
                    {
                        dp[i|(1<<j)]=max(dp[i|(1<<j)],dp[i]+g[k][j]);
                        //dp[i|(1<<k)]=max(dp[i|(1<<k)],dp[i]+g[j][k]);
                    }
        }
        printf("%d\n",ans);
    }
    return 0;
}