ZOJ 3471 Most Powerful

Most Powerful

Time Limit: 2000ms

Memory Limit: 65536KB

This problem will be judged on ZJU. Original ID: 3471
64-bit integer IO format: %lld      Java class name: Main

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, ... , AN. 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 Ai and Aj collide with Aj 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

 

Source

ZOJ Monthly, February 2011

Author

GAO, Yuan

 

解题：状压dp

 

用二进制1表示剩下的原子

 

#include <bits/stdc++.h>
using namespace std;
const int maxn = 12;
int dp[1<<maxn],power[maxn][maxn],n;
int main(){
    while(scanf("%d",&n),n){
        for(int i = 0; i < n; ++i)
            for(int j = 0; j < n; ++j)
                scanf("%d",power[i]+j);
        memset(dp,0,sizeof dp);
        for(int i = (1<<n)-1; i > 0; --i){
            for(int j = 0; j < n; ++j){
                if(i&(1<<j)){
                    for(int k = 0; k < n; ++k){
                        if((i&(1<<k)) == 0 || j == k) continue;
                        dp[i^(1<<k)] = max(dp[i^(1<<k)],dp[i]+power[j][k]);
                    }
                }
            }
        }
        int ret = 0;
        for(int i = 0; i < n; ++i) ret = max(ret,dp[1<<i]);
        printf("%d
",ret);
    }
    return 0;
}