POJ-1321 board Problem n queen problem (Binary Version)

POJ-1321 board Problem n queen problem (Binary Version)

Give an n * n board, which cannot contain pawns in some places.
Now you are asked to place k pieces on the board so that any two of these pieces are not in the same column. How many methods are there?

Solution: similar to the problem of Queen n, it can be processed in one row. The value 1 in binary indicates that the location is retainable, and the value 0 indicates that the location is not retainable, all indicates the status where each row can be put (1 <n)-1)
If the column state is s, s & (restriction of the row) & All is the final state of the row.
Now we only need to find all the 1 in this final state, so how can we quickly find this 1
If the final state is ss, then ss & (-ss) & All is the first 1 position (which can be verified by yourself ), set a while loop to find All 1. Remember & All

#include
  
   #includeusing namespace std;const int N = 20;char str[N];int n, k, ans, All, statu[N];void init() {    All = (1 << n) - 1;    for(int i = 0; i < n; i++) {        gets(str);        statu[i] = All;        for(int j = 0; j < n; j++)            if(str[j] == '.')                statu[i] ^= (1 << j);    }    ans = 0;}void solve(int cur, int num, int s) {    if(num == k) {        ans++;        return ;    }    if(cur == n || n - cur + num < k)         return ;    int ss = (s & statu[cur]) & All;     while(ss) {        int t = ss & (-ss) & All;        solve(cur + 1, num + 1, s ^ t);        ss = (ss ^ t )& All;    }    solve(cur + 1, num, s);}int main() {    while(gets(str)) {        sscanf(str,"%d%d", &n, &k);        if(n == -1 && k == -1)            break;        init();        solve(0,0,All);        printf("%d\n", ans);    }    return 0;}