POJ 3600 Subimage Recognition 搜索

Subimage Recognition

Time Limit: 1000MS Memory Limit: 131072KTotal Submissions: 3072 Accepted: 1176

Description

An image A is said to be a subimage of another image B if it is possible to remove some rows and/or columns of pixels from B so that the resulting image is identical to A. Figure 6 illustrates an example. Image A, shown in Figure 6(a), is a subimage of image B, shown in Figure 6(b), because the image resulting from the removal of the middle row and column of pixels from B is identical to A.

            (a) Image A (b) Image B

Figure 6: An example of a subimage

Given two black-and-white images A and B, determine whether A is a subimage of B.

Input

The input contains a single test case. The first line contains two integers r and c (1 ≤ r, c ≤ 20), the dimensions of A. The following r lines, each containing a string of length c, give an r × c 0-1 matrix representing the pixels of A. The next line contains two integers R and C (r ≤ R ≤ 20; c ≤ C ≤ 20), the dimensions of B. The following R lines, each containing a string of length C, give an R × C 0-1 matrix representing the pixels of B. A 0 indicates a white pixel; a 1 indicates a black pixel.

Output

If A is a subimage of B, print “Yes”; otherwise, print “No”.

Sample Input

2 211103 3101000100

Sample Output

Yes

题意：给一个01矩阵A，给一个01矩阵B，问删除B的某些行和B的某些列能否使得矩阵A和矩阵B一模一样。

解法：删除一些行和列使得和A相等，也可以理解成在B中是否存在和A相同的子矩阵（元素不一定，行列的相对坐标对齐就行）。我的方法是在B的每一行中搜索与A第一行相同的所有方案，对可行方案的列进行标记。对于每一种方案，从开始标记的那一行开始往下逐行判断，能够成一个A矩阵就yes，不然就继续搜。

#include <stdio.h>#include <iostream>#include <algorithm>#include <string.h>using namespace std;const int N = 100;int an,am,bn,bm;char A[N][N];char B[N][N];bool vis[N];  ///标记合法方案的列bool flag;    ///标记是否能构成A矩阵void judge()  ///对于B标记的列，每行与A进行匹配{    int ax = 1,ay = 1;    for(int bx = 1;bx <= bn;bx++){        bool lead = true;        for(int i = 1;i <= bm;i++){            if(vis[i]){                if(A[ax][ay++] != B[bx][i]){                    lead = false;                    break;                }            }        }        if(lead) ax++;  ///成功匹配一行，就匹配下一行        ay = 1;        if(ax > an) break;    }    if(ax > an) flag = true;}void dfs(int a,int b,int c)  ///A中第一行的第a个元素，B中c行的第b个元素{    if(flag) return;    if(a == am+1){        judge();        return;    }    if(b > bm) return;    for(int i = b;i <= bm;i++){        if(A[1][a] == B[c][i]){            vis[i] = true;            dfs(a+1,i+1,c);            vis[i] = false;        }    }}int main(void){    scanf("%d%d",&an,&am);    flag = false;    memset(vis,false,sizeof vis);    for(int i = 1;i <= an;i++)        scanf("%s",A[i]+1);    scanf("%d%d",&bn,&bm);    for(int i = 1;i <= bn;i++)        scanf("%s",B[i]+1);    for(int i = 1;i <= bn;i++){        for(int j = 1;j <= bm;j++){            if(A[1][1] == B[i][j]){                dfs(1,j,i);            }        }    }    if(flag) puts("Yes");    else puts("No");    return 0;}