1019 Most Powerful 状压DP-NP问题

链接：https://ac.nowcoder.com/acm/contest/25022/1019
来源：牛客网

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

复制 2 0 4 1 0 3 0 20 1 12 0 1 1 10 0 0

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

输出

复制 4 22

4
22

 

分析

每两种气体相撞都有不同情况，所以是np问题。

设f[i] 表示状态i 最大的能量

第k位为0，表示第k中气体还没消失，为1表示消失了。

f[i | (1 << (k - 1) ] = max(f[i | ( 1 << (k - 1) + a[j][k] ) 表示用第j 种气体去撞第k种气体

//-------------------------代码----------------------------

//#define int ll
const int N = 11;
int n,m;

int f[1<<11];
int a[N][N];;

void solve()
{
    ms(f,0);
    fo(i,1,n) {
        fo(j,1,n) {
            cin>>a[i][j];
        }
    }
    fo(i,0,(1<<n) - 1) {
        fo(j,1,n) {
            if(i & (1 << (j - 1))) continue;
            fo(k,1,n) {
                if(i & (1 << (k - 1) )) continue;
                f[i| (1 <<(k-1))] = max(f[i | ( 1 << ( k - 1 ) )],f[i] + a[j][k]);
            }
        }
    }
    cout<<f[(1 << n ) - 1] <<endl;
}

signed main(){
    AC();
    clapping();TLE;

//    int t;cin>>t;while(t -- )
    while(cin>>n,n)
    solve();
//    {solve(); }
    return 0;
}

/*样例区


*/

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