zoj3471

Most Powerful

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Time Limit: 2 Seconds Memory Limit: 65536 KB

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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：给出n个原子，a和b碰撞会产生w[a][b]的能量且b会消失（总感觉题目这里没说清楚，哪颗会消失），剩下的那颗能去撞别的也能被别的撞，问最后总的能量最多是多少。

思路：状压dp，用1表示其存在，0表示消失，只要从全部都存在枚举到全部都消失即可。dp[i^(1<<k)]=max(dp[i^(1<<k)],dp[i]+w[j][k]); 最后取剩某一颗的最大值

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int w[15][15];int dp[1<<15];int main(void){int n;while(scanf("%d",&n),n){for(int i=0;i<n;i++)for(int j=0;j<n;j++)scanf("%d",&w[i][j]);memset(dp,0,sizeof(dp));for(int i=(1<<n)-1;i>0;i--)for(int j=0;j<n;j++)if((1<<j)&i)for(int k=0;k<n;k++)if(k!=j&&((1<<k)&i))dp[i^(1<<k)]=max(dp[i^(1<<k)],dp[i]+w[j][k]);int ans=0;for(int i=0;i<n;i++)ans=max(ans,dp[1<<i]);printf("%d\n",ans);}return 0;}