Most Powerful（状压dp）

Most Powerful

 

题目描述

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. 

输入描述:

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

The last case isfollowed by a 0 in one line.

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

输出描述:

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

示例1

输入

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

输出

4
22

题目思路
dp[i][j]表示进行i此操作后所有原子状态为j的最大值,j中第k位为1表示第k个原子已消失。
最后的答案要循环一遍枚举（(1<位运算不清楚优先级的地方一定要记得打括号，这里卡了我好久。。。
代码如下

#include
#define ll long long
using namespace std;
int n,dp[11][1<<10],m[11][11],ans;
int main()
{
    while(1)
    {
        ans = 0;
        memset(dp,0,sizeof(dp));
        cin>>n;
        if(n==0)
            break;
        for(int i = 0;i)
            for(int j = 0;j)
                scanf("%d",&m[i][j]);
        for(int i = 1;i//阶段 
            for(int j = 0;j<1<//状态
                for(int k = 0;k)
                    for(int h = 0;h)
                        if(k!=h&&(!(j>>k&1))&&(!(j>>h&1)))
                            dp[i][j|(1<1<1][j]+m[h][k]);
        for(int i = 0;i)
            ans = max(ans,dp[n-1][(((1<1)&(~(1<<i)))]);
        cout<endl;
    }
    return 0;
}