Most Powerful - ZOJ 3471 状压dp

原子碰撞最大能量计算

最新推荐文章于 2021-08-17 01:05:19 发布

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本文介绍了一道关于原子碰撞产生能量的问题，通过状态压缩动态规划（状压DP）的方法求解如何让N个不同的原子相互碰撞以产生最大的总能量。

Most Powerful

Time Limit: 2 Seconds Memory Limit: 65536 KB

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.

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

题意：两个原子弹碰到一起后会产生一定的能量，并且其中一个会消失，问你最多产生多少能量。

思路：状压dp，dp[S]表示在S状态下能够产生的最多的能量。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[4000],num[15][15];
int main()
{ int n,i,j,k,S,S2,ans;
  while(~scanf("%d",&n)&& n>0)
  { for(i=0;i<n;i++)
     for(j=0;j<n;j++)
      scanf("%d",&num[i][j]);
    memset(dp,0,sizeof(dp));
    S2=1<<n;
    for(S=0;S<S2;S++)
     for(i=0;i<n;i++)
     { if(S&(1<<i))
        continue;
       for(j=0;j<n;j++)
       { if(i==j || (S&(1<<j)))
          continue;
          dp[S+(1<<j)]=max(dp[S+(1<<j)],dp[S]+num[i][j]);
       }
     }
    ans=0;
    for(i=0;i<n;i++)
     ans=max(ans,dp[S2-1-(1<<i)]);
    printf("%d\n",ans);
  }
}