+ Binary enumeration thinking off white cow May 23 A title race membrane record

Record membrane

Title Description

Recently playing a beef called "membrane record" of the game, the mechanism of this game is this:
in a game, all of the enemies are arranged in a grid of n rows and m columns, Taurus command of his magical girl to attack the enemy.
Attack There are two types: Line blast, columns blast
line blast can destroy the enemy a whole row, column blast can destroy the enemy a whole column of
beef in total capable of releasing a trip to blast, b minor columns blast
given the initial situation of a bureau of the game I ask whether beef will wipe out the enemy?

Enter a description:

The first line contains a positive integer T, denotes the number of the test data set, followed by a set of test data T
each test first row of data has four positive integers n, m, a, b
Then there are n rows, each row m is a string of characters of the i-th row j-th column is illustrated here * If there is an enemy, if Description there is no

Output Description:

For each set of test data output line, if they can destroy all the enemies on the output yes, otherwise no output

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Let me talk about, solving the problem of data may be some problems; to consider a = 0;

The very thought of a question;

First thought may hold binary pieces, the i-th pre-fetch a binary number (i.e., a combination of lines), number of columns to eliminate these lines are eliminated, referred to as CNT [i]; the number of columns and m is the rest is -cnt [i];

How is the key to solving the problem preprocessing, first column from the enumeration of 1-m, and the position of the enemy lines where each column is processed, passed into a binary number; each binary number in this column line enemy arrangement; then CNT [i] represents the number of enemies are arranged in columns i;

There are treatments, such as: 11011 arrangement, we must take the equivalent of 1,2,4,5-line before they can destroy the enemy, but when we take 1.2.4.5 also put 11010,11000, etc. are eliminated, so it to find the number of permutations of a binary number 1 weight each. You should be able to understand this myself;

Code:

#include<bits/stdc++.h>
#define LL long long
#define pa pair<int,int>
#define ls k<<1
#define rs k<<1|1
#define inf 0x3f3f3f3f
using namespace std;
const int N=100100;
const int M=2000100;
const LL mod=100000000;
int n,m,a,b,cnt[M];
char s[25][N];
int sum_1(int p){
    int ans=0;
    while(p){
        if(p%2) ans++;
        p/=2;
    }
    return ans;
}
int main(){
    int t;
    cin>>t;
    while(t--){
        cin>>n>>m>>a>>b;
        for(int i=1;i<=n;i++) scanf("%s",s[i]+1);
        for (int i = 0; i < (1<<n); i++) cnt[i] = 0;
        for(int j=1;j<=m;j++){
            int t=0;
            for(int i=1;i<=n;i++){
                if(s[i][j]=='*') t|=(1<<i-1);//t代表第i列有哪几行有敌人
            }
            cnt[t]++;//相同列的个数
        }
        for(int i=1;i<=n;i++){
            for(int j=0;j<=(1<<n)-1;j++){
                if((j&(1<<i-1))==0){//这个表示数j的第i个二进制位为0
                    cnt[j|(1<<i-1)]+=cnt[j];//表示第i个二进制位为1的数加上cnt[j]
                }
            }
        }
        bool ok=false;
        for(int i=0;i<(1<<n);i++){
            if(sum_1(i)<=a&&m-cnt[i]<=b) ok=true;
        }
        if(ok) cout<<"yes"<<endl;
        else cout<<"no"<<endl;
    }	
    return 0;
}