zoj 3471(状态压缩DP,类似于点集配对)

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

解法1：

所有的原子组成一个集合，每次从中选取两个点，选择一个攻击点，选择一个被攻击点，

d[s]=max(d[s],d[s ^ (1<<j)] +a[i][j]);

#include <iostream>

#include <cstdio>

#include <cstring>

#include <cstdlib>

#define maxn 10

using namespace std;

int d[<<]; //表示到达状态s时产生的最大能量

int a[][];

int n;

void init()

{

   memset(d,,sizeof(d));

}

void solve()

{

   for(int s=;s<(<<n);s++)

   {

       d[s]=;

      for(int i=;i<n;i++)

       if(s & (<<i))

       {

           for(int j=;j<n;j++)

            if(s & (<<j))

           {

               if(i==j)

                continue;

               d[s]=max(d[s],d[s ^ (<<j)]+a[i][j]);

           }

       }

   }

}

int main()

{

    while(~scanf("%d",&n))

    {

      if(n==)

        break;

        init();

      for(int i=;i<n;i++)

        for(int j=;j<n;j++)

      {

          scanf("%d",&a[i][j]);

      }

      solve();

      int ans=;

    // for(int i=0;i<(1<<n);i++)

     //   cout<<d[i]<<" ";

     //cout<<endl;

      //ans=max(ans,d[i]);

     printf("%d\n",d[(<<n)-]);

    }

    return ;

}

解法2：

假设一个数，第i位表示第i个原子是否被灭掉，如果被灭掉则为1，没被灭掉为0，那么所有状态都可以用2^n范围内的数来表示。则初始状态为0，即所有原子都没有消失

　　令dp[i]表示达到状态 i 时所产生的最大能量，则答案就是从0~（1<<n）所有状态里释放的最大的那个能量。需要枚举所有状态。

　　假设当前状态是s，从1~n里边枚举主动碰撞的原子 i ，和被动碰撞被消灭掉的原子 j ，则

　　dp[s | (1<<j)] = max{dp[s | (1<<j)] , dp[s] + A[i][j]};

#include <iostream>

#include <cstdio>

#include <cstring>

#include <cstdlib>

#define maxn 10

using namespace std;

int d[<<]; //表示到达状态s时产生的最大能量

int a[][];

int n;

void init()

{

    memset(d,,sizeof(d));

}

void solve()

{

    for(int s=;s<(<<n);s++)

    {

     for(int i=;i<n;i++)

     {

       if( ! (s & (<<i)) ) //不知为何写成(if( (s & (1 << i)) ))也能过

       {

            for(int j=;j<n;j++)

          {

           if(i==j)

          continue;

           if(s & (<<j))

           continue;

          d[s |(<<j)] = max(d[s | (<<j)],d[s]+a[i][j]);

         }

      }

     }

    }

}

int main()

{

    while(~scanf("%d",&n))

    {

        if(n==)

        break;

        init();

        for(int i=;i<n;i++)

        for(int j=;j<n;j++)

        {

           scanf("%d",&a[i][j]);

        }

        solve();

        int ans=;

        for(int i=;i<(<<n);i++)

        ans=max(ans,d[i]);

        printf("%d\n",ans);

    }

    return ;

}

巴特西

zoj 3471(状态压缩DP,类似于点集配对)

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