Most Powerful(状压dp)
Most Powerful
题目描述
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.输入描述:
There are multiplecases. The first line of each case has an integer N (2 <= N <= 10), whichmeans there are N atoms: A1, A2, ... , AN.Then N lines follow. There are N integers in each line. The j-th integer on thei-th line is the power produced when Ai and Aj collidewith Aj gone. All integers are positive and not larger than 10000.
The last case isfollowed by a 0 in one line.
There will be no morethan 500 cases including no more than 50 large cases that N is 10.
输出描述:
Output the maximalpower these N atoms can produce in a line for each case.
输入
2 0 4 1 0 3 0 20 1 12 0 1 1 10 0 0
输出
4 22
题目思路
dp[i][j]表示进行i此操作后所有原子状态为j的最大值,j中第k位为1表示第k个原子已消失。
最后的答案要循环一遍枚举((1<<n)-1)状态每一位赋值0的最大值,因为n个原子只能有n-1次操作。
位运算不清楚优先级的地方一定要记得打括号,这里卡了我好久。。。
代码如下
#include<bits/stdc++.h> #define ll long long using namespace std; int n,dp[11][1<<10],m[11][11],ans; int main() { while(1) { ans = 0; memset(dp,0,sizeof(dp)); cin>>n; if(n==0) break; for(int i = 0;i<n;i++) for(int j = 0;j<n;j++) scanf("%d",&m[i][j]); for(int i = 1;i<n;i++)//阶段 for(int j = 0;j<1<<n;j++)//状态 for(int k = 0;k<n;k++) for(int h = 0;h<n;h++) if(k!=h&&(!(j>>k&1))&&(!(j>>h&1))) dp[i][j|(1<<k)] = max(dp[i][j|(1<<k)],dp[i-1][j]+m[h][k]); for(int i = 0;i<n;i++) ans = max(ans,dp[n-1][(((1<<n)-1)&(~(1<<i)))]); cout<<ans<<endl; } return 0; }