ZOJ----3471Most powerful（简单状压dp）

题目

Most Powerful
Time Limit: 2 Seconds Memory Limit: 65536 KB
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

大意：有很多种气体（n<=10），任意两种气体可以相互反应，消耗掉其中一种气体并释放出一定的能量（消耗两种气体释放的能量不一样），求最大释放的能量有多组数据

随便写写的水题，dp[state]表示当前状态为state时的最大消耗，枚举还没被消耗的气体就可以转移状态了

#include<bits/stdc++.h>
using namespace std;
int dp[2000],n,p[12][12];
int main(){
    	while(scanf("%d",&n),n)
    	{
        		memset(dp,0,sizeof(dp));
        		for(int i=1;i<=n;i++)
        		{
    	        		for(int j=1;j<=n;j++)
	            		{
	                			scanf("%d",&p[i][j]);
	            		}
	        	}
	        	int state=1<<n;
	        	for(int i=0;i<state;i++)
	    	    {
	        	    	for(int j=1;j<=n;j++)
	        		    {
		                		if(i&(1<<(j-1))) continue;
	                			for(int k=1;k<=n;k++)
            	    			{
		                    			if(j==k) continue;
		                    			if(i&(1<<(k-1))) continue;
                    					dp[i+(1<<(k-1))]=max(dp[i+(1<<(k-1))],dp[i]+p[j][k]);
                				}
            			}
	        	}
    	    	int ans=0;
	        	for(int i=0;i<state;i++)
	        	{
	            		ans=max(ans,dp[i]);
        		}
        		printf("%d
",ans);
        	}
    	return 0;
}