zoj3471 Most Powerful

最新推荐文章于 2019-09-27 19:59:13 发布

Kirito_Acmer

最新推荐文章于 2019-09-27 19:59:13 发布

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分类专栏：状压dp 文章标签：状压dp

本文链接：https://blog.youkuaiyun.com/Kirito_Acmer/article/details/46609861

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本文探讨了在给定的原子集合中，通过特定的碰撞规则，如何实现原子碰撞产生的能量最大化的问题。通过状态压缩的方法，利用动态规划解决原子碰撞序列优化，最终确定剩余原子能够产生的最大能量。

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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

题意是有n个气球，n的范围是1~n，每两个气球碰撞都会产生一定的能量，并且有一个气球会消失，问最后剩下一个气球的时候最多能产生多少能量。可以用状态压缩，设0表示气体存在，1表示气体消失，状态转移方程dp[state]=max{dp[state],dp[state']+a[i][j]}.

#include<stdio.h>
#include<string.h>
int a[15][15],dp[1500];
int max(int a,int b){
	return a>b?a:b;
}
int main()
{
	int n,m,i,j,s,ans;
	while(scanf("%d",&n)!=EOF && n!=0)
	{
		for(i=1;i<=n;i++){
			for(j=1;j<=n;j++){
				scanf("%d",&a[i][j]);
			}
		}
		memset(dp,0,sizeof(dp));
		for(s=1;s<(1<<n);s++){
			for(j=1;j<=n;j++){
				if(s&(1<<(j-1))){
					for(i=1;i<=n;i++){
						if( !(s&(1<<(i-1))) ){
							dp[s]=max(dp[s],dp[s-(1<<(j-1))]+a[i][j]);
						}
					}
				}
			}
		}
		ans=0;
		for(i=1;i<=n;i++){
			if(dp[(1<<n)-1-(1<<(i-1))]>ans)ans=dp[(1<<n)-1-(1<<(i-1))];
		}
		printf("%d\n",ans);
	}
	return 0;
}