spoj-ASSIGN-bitDP

ASSIGN - Assignments

#dynamic-programming

 

Problem

Your task will be to calculate number of different assignments of n different topics to n students such that everybody gets exactly one topic he likes.

Input

First line of input contains number of test cases c (1<=c<=80). Each test case begins with number of students n (1<=n<=20). Each of the next n lines contains n integers describing preferences of one student. 1 at the ith position means that this student likes ith topic, 0 means that he definitely doesn't want to take it.

Output

For each test case output number of different assignments (it will fit in a signed 64-bit integer).

Example

Input:
3
3
1 1 1
1 1 1
1 1 1
11
1 0 0 1 0 0 0 0 0 1 1 
1 1 1 1 1 0 1 0 1 0 0 
1 0 0 1 0 0 1 1 0 1 0 
1 0 1 1 1 0 1 1 0 1 1 
0 1 1 1 0 1 0 0 1 1 1 
1 1 1 0 0 1 0 0 0 0 0 
0 0 0 0 1 0 1 0 0 0 1 
1 0 1 1 0 0 0 0 0 0 1 
0 0 1 0 1 1 0 0 0 1 1 
1 1 1 0 0 0 1 0 1 0 1 
1 0 0 0 1 1 1 1 0 0 0 
11
0 1 1 1 0 1 0 0 0 1 0 
0 0 1 1 1 1 1 1 1 1 1 
1 1 0 1 0 0 0 0 0 1 0 
0 1 0 1 0 1 0 1 0 1 1 
1 0 0 1 0 0 0 0 1 0 1 
0 0 1 0 1 1 0 0 0 0 1 
1 0 1 0 1 1 1 0 1 1 0 
1 0 1 1 0 1 1 0 0 1 0 
0 0 1 1 0 1 1 1 1 1 1 
0 1 0 0 0 0 0 0 0 1 1 
0 1 1 0 0 0 0 0 1 0 1 

Output:
6
7588
7426

 Submit solution!

 

                          利用二进制表示集合，f[S][i]表示i个人形成S状态的座位的方案个数，答案就是 f[all][n],  注意爆int。

     

#include<bits/stdc++.h>
using namespace std;
#define inf 0x3f3f3f3f
long long  f[2][(1<<20)+10];
int e[21][21];
int main()
{
    int t,n,m,i,j,k;
    cin>>t;
    while(t--){
        cin>>n;
        for(i=1;i<=n;++i) 
        for(j=1;j<=n;++j) cin>>e[i][j];
        memset(f,0,sizeof(f));
        f[0][0]=1;
        int cur=0;
        for(i=1;i<=n;++i){
            cur^=1;
            memset(f[cur],0,sizeof(f[cur]));
            for(j=0;j<(1<<n);++j){
                if(!f[cur^1][j]) continue;
                for(k=1;k<=n;++k){
                if(e[i][k]&& ((1<<(k-1))&j)==0) {
                    f[cur][j|(1<<(k-1))]+=f[cur^1][j];
                }
            }
            }
            /*for(j=0;j<(1<<n);++j){
                cout<<j<<' '<<f[cur][j]<<endl;
            }*/
        }
        cout<<f[cur][(1<<n)-1]<<endl;
    }
    return 0;
}

 

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