SCU-3088 windy's cake VIII

状态压缩DP

Description

After wqb0039 has made the n*n cake.
We insert candles in the cake.
But we hate the square,We change the cake to n*m size by do some grids changes.
Now we have a cake of n*m size,and we want to insert k candles in the cake.
A grid of the cake can only take one or no candle.
Moreover no two candles can in the same row or same column.
How many ways can we put all the k candles into the n*m cake.

Input

Multiple test cases.
Each contains n+2 line.
Line 1: Contains three integers n m and k.
Line 2 ~ n+1: Each contains m integers which is 1 or 0.
1 means the gird can take one candle.
0 means the gird can take no candle.
Line n+2: The blank line.

( 1 <= n,m,k <= 20 )

Output

For each test case,output only one line,the answers mod 19871118.

Sample Input

1 1 1
1

1 1 1
0

2 2 2
1 1
1 1

2 2 2
1 1
1 0

4 3 2
0 0 1
0 1 1
1 1 0
1 0 0

4 3 3
0 0 1
0 1 1
1 1 0
1 0 0

2 2 3
1 1
1 1

10 10 5
0 1 0 0 0 0 1 0 0 0
0 0 1 1 0 1 0 0 0 0
1 0 1 1 0 1 0 0 1 1
0 0 1 1 1 1 0 1 0 0
1 1 1 1 1 1 0 1 1 1
1 1 0 1 1 0 0 1 0 0
0 0 0 1 0 0 1 1 1 0
0 1 1 0 0 1 1 1 1 0
1 1 1 1 0 1 1 1 1 1
0 1 1 0 0 1 1 0 1 1

Sample Output

1
0
2
1
10
4
0
323825

题目大意：
插蜡烛，不能有两个蜡烛在同一行或者在同一列，而且只能在给定的图的标识为1的地方插入蜡烛，问插入k个蜡烛一共有多少插法。

话说这个题我拿到首先得想法是暴力回溯，结果弄了老久还没想明白。后来还是参考别人的思想，用Dp来做，作为蒟蒻整整搞了半天才想明白。

计算二进制数中1位的方法很不错

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

int d[22][(1 << 20) | 1];
int cnt[(1 << 20) | 1];
int a[22];
int m,n,k;
const int mod = 19871118;
void init()//此处是用来计算二进制数中一的个数，这种方法对我来说有点不好理解
{           //但是直接计算的话又会超时，最后还是搞明白了
    for(int i = 1; i < (1<<20); i++)
    {
 //i&(-i)的结果是i最右边的那个1的位是1，其他全是0，就是说0110做运算结果是0010
        cnt[i]=cnt[i - (i &(-i)) ]+1;
    }
}
int read()
{
    int s;
    if((s=scanf("%d %d %d",&n,&m,&k))!=EOF)
    {
        memset(d, 0, sizeof(d));
        memset(a, 0, sizeof(a));
        for(int i = 1; i <= n; i++)
        {
            for(int j =0; j < m; j++)
            {
                int x;
                scanf("%d",&x);
                if(x) a[i]=a[i]|(1 << j);//状态压缩，以二进制储存图的信息
            }
        }
    }
    return s;
}
void dp()
{
    d[0][0]=1;
    for(int i = 0; i <= n; i++)
    {
        for(int j = 0; j < (1<<m); j++)
        {
            int x =(j & a[i+1]) ^ a[i+1];//枚举状态j,得出x是1表示可以插但没有插的状态
            for(int s = x; s > 0; s -= s&(-s) )
            {
                int o =s & (-s);//依次处理插入位置的状态dp，每行只能插一个
                d[i+1][j | o]+=d[i][j];
                d[i+1][j | o]%=mod;
            }
            d[i+1][j]+=d[i][j];//更新一个都没有插的状态
            d[i+1][j]%=mod;
        }
    }
}

int main()
{
    init();
    //freopen("D://input.txt","r",stdin);
    while(read()!=EOF)
    {
        dp();
        int ans=0;
        for(int i = 0; i < (1<<m); i++)
            if(cnt[i]==k) ans=(ans+d[n][i])%mod;
        printf("%d\n",ans);
    }
    return 0;
}