ZOJ 3471 Most Powerful (状压dp)

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

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Author: GAO, Yuan

Contest: ZOJ Monthly, February 2011

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3471

题目大意：n个粒子，ai和aj碰撞会产生aij能量同时aj消失，求最终能产生的最大能量

题目分析：看数据量后很明显的状压，dp[i]表示状态为i时的最大能量，状态为:第i位为1则表示第i-1个粒子已经被使用，否则未使用，然后枚举状态，选择两个没有被使用的不同的粒子尝试碰撞

dp方程为dp[i | (1 << k)] = max(dp[i | (1 << k)], dp[i] + a[j][k])，其中i表示状态，k表示碰撞的后一个粒子，j表示碰撞的前一个粒子

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int n, a[15][15];int dp[1 << 11];int main(){while(scanf("%d", &n) != EOF && n){for(int i = 0; i < n; i++)for(int j = 0; j < n; j++)scanf("%d", &a[i][j]);memset(dp, 0, sizeof(dp));for(int i = 0; i < (1 << n); i++)for(int j = 0; j < n; j++)if(!(i & (1 << j)))for(int k = 0; k < n; k++)if(j != k && !(i & (1 << k)))dp[i | (1 << k)] = max(dp[i | (1 << k)], dp[i] + a[j][k]);int ans = 0;for(int i = 0; i < (1 << n); i++)ans = max(ans, dp[i]);printf("%d\n", ans);}}