Most Powerful（状压dp入门）-优快云博客

本文链接：https://blog.youkuaiyun.com/x_Armin/article/details/98483872

描述

传送门：ZOJ-3471

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: $A_1, A_2, … , A_N$. 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.

Output

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

Examples

intput
1
2
3
4
5
6
7
8
2
0 4
1 0
3
0 20 1
12 0 1
1 10 0
0
output
1
2
4
22

思路

状态dp入门题，用0来表示未泯灭，1表示已经泯灭，这样就能用一个n位的整数来记录所有的状态。
对于状态i，假设第j位和第k位都是0，那么我们就可以通过dp[i]推出 dp[i&1<<k]的状态。
状态转移方程：

代码

#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for(int i=a;i<n;i++)
#define repd(i,a,n) for(int i=n-1;i>=a;i--)
#define CRL(a,x) memset(a,x,sizeof(a))

int book[12][12],n,dp[1<<12],ans=0;

int main()
{
    while(scanf("%d",&n)&&n){
        CRL(dp,0);ans=0;
        rep(i,0,n)
            rep(j,0,n)
                scanf("%d",&book[i][j]);

        rep(i,0,1<<n)       //所有状态
            rep(j,0,n)      //泯灭的原子
                if(!(i&1<<j))//泯灭的原子要是0
                    rep(k,0,n)//被碰撞的原子
                        if(j!=k&&!(i&1<<k)) //被碰撞的原子要是0且不是泯灭的原子
                            dp[i|1<<k]=max(dp[i]+book[j][k],dp[i|1<<k]);

        rep(i,0,1<<n) ans=max(dp[i],ans);
        printf("%d\n",ans);
    }
    return 0;
}