【状压dp】Most Powerful

[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

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Author: GAO, Yuan
Contest: ZOJ Monthly, February 2011

题目大意：给你一些元素，这些元素碰撞后会产生其中消失的物品的能量(你可以自己决定哪一个消失)，求最后剩1个元素的最大能量释放值

试题分析：比较基础的一道状压dp      貌似在dp中我只能说状压dp有基础题，好弱QAQ

　　　　　dp[S]表示序列消到状态为S的时候的最大释放能量值。

　　　　　可得出如下转移方程(因为是消失，所以应该从(1<<N)-1到1枚举)： dp[S]=max(dp[S],dp[S+(1<<(k-1))]+make[j][k]);

　　　　　其中k代表消失的能量，j代表剩下的能量，前提是S包含j，不包含k，k!=j

 

代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<stack>
#include<algorithm>
using namespace std;

inline int read(){
	int x=0,f=1;char c=getchar();
	for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
	for(;isdigit(c);c=getchar()) x=x*10+c-'0';
	return x*f;
}
const int MAXN=100001;
const int INF=999999;
int N,M;
int dp[MAXN];
int e[1001][1010];

int main(){
    N=read();
    while(1){
    	if(!N) break;
    	memset(dp,0,sizeof(dp));
    	for(int i=1;i<=N;i++){
    		for(int j=1;j<=N;j++)
    		    e[i][j]=read();
		}
		int ans=0;
		for(int i=(1<<N)-2;i>=1;i--){
			for(int j=1;j<=N;j++){
				if(!((i>>(j-1))&1)) continue;
				for(int k=1;k<=N;k++){
					if(k==j||((i>>(k-1))&1)) continue;
					dp[i]=max(dp[i],dp[i+(1<<(k-1))]+e[j][k]);
				}
			}
		}
		for(int i=1;i<=N;i++) ans=max(ans,dp[1<<(i-1)]);
		printf("%d
",ans);
		N=read();
	}
}