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[][(<<)+];
 int e[][];
 int main()
 {
     int t,n,m,i,j,k;
     cin>>t;
     while(t--){
         cin>>n;
         for(i=;i<=n;++i)
         for(j=;j<=n;++j) cin>>e[i][j];
         memset(f,,sizeof(f));
         f[][]=;
         int cur=;
         for(i=;i<=n;++i){
             cur^=;
             memset(f[cur],,sizeof(f[cur]));
             for(j=;j<(<<n);++j){
                 if(!f[cur^][j]) continue;
                 for(k=;k<=n;++k){
                 if(e[i][k]&& ((<<(k-))&j)==) {
                     f[cur][j|(<<(k-))]+=f[cur^][j];
                 }
             }
             }
             /*for(j=0;j<(1<<n);++j){
                 cout<<j<<' '<<f[cur][j]<<endl;
             }*/
         }
         cout<<f[cur][(<<n)-]<<endl;
     }
     return ;
 }

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