zoj3471 一维状压dp
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
此题比较简单，我们dp[s]表示状态为s时的最大能量，注意这里反面思考，0表示表示原子存在，1表示原子不存在，这样初态就是dp[0]=0,然后继续用刷表法。
代码：
#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int adj[12][12],dp[1024];
int main()
{
int n;
while(cin>>n,n){
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
cin>>adj[i][j];
memset(dp,0,sizeof dp);
for(int s=0;s<1<<n;s++){
for(int i=0;i<n;i++){
if(s&1<<i) continue; //i气体不存在
for(int j=0;j<n;j++){
if(i==j||s&1<<j) continue; //同一种气体或者j气体不存在
int ns=s+(1<<j); //j气体消失
dp[ns]=max(dp[ns],dp[s]+adj[i][j]);
}
}
}
int ans=0;
for(int s=0;s<1<<n;s++)
ans=max(dp[s],ans);
cout<<ans<<endl;
}
return 0;
}