DP状态压缩：Most Powerful

最新推荐文章于 2017-12-10 20:27:35 发布

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本文介绍了一种通过编程计算N个不同原子碰撞所能产生的最大总能量的方法。在每次碰撞中，两个原子相撞会产生一定数量的能量，并且其中一个原子会消失。文章提供了一个具体的程序实现示例，展示了如何通过算法来找出所有可能碰撞中最优的能量产出方案。

Most Powerful Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu

Submit Status

Description

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

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>

using namespace std;

int p[15][15];
int ps[1<<11];

int main(){
    int n, i, j;
    while(cin >> n && n){
        for(i = 0; i < n; i ++){
            for(j = 0; j < n; j ++){
                scanf("%d", &p[i][j]);
            }
        }
        int allS = 1 << n;
        int curS; 
        memset(ps, 0, sizeof(ps));
        for(curS = 0; curS < allS; curS ++){
            for(i = 0; i < n; i ++){
                if(curS & (1 << i))
                    continue;
                for(j = 0; j < n; j ++){
                    if(j == i)
                        continue;
                    if(curS & (1 << j))
                        continue;
                    int newS = curS + (1<<j);
                    ps[newS] = max(ps[newS], ps[curS] + p[i][j]);
                }
            }
        }
        int ans = 0;
        for(i = 0; i < allS; i ++){
            if(ps[i] > ans)
                ans = ps[i];
        }
        printf("%d\n", ans);
    }
    return 0;
}