CF14A Letter（模拟+字符串）（C++题解）（大佬勿喷）
题目描述
A boy Bob likes to draw. Not long ago he bought a rectangular graph (checked) sheet with nn rows and mm columns. Bob shaded some of the squares on the sheet. Having seen his masterpiece, he decided to share it with his elder brother, who lives in Flatland. Now Bob has to send his picture by post, but because of the world economic crisis and high oil prices, he wants to send his creation, but to spend as little money as possible. For each sent square of paper (no matter whether it is shaded or not) Bob has to pay 3.14 burles. Please, help Bob cut out of his masterpiece a rectangle of the minimum cost, that will contain all the shaded squares. The rectangle's sides should be parallel to the sheet's sides.
输入格式
The first line of the input data contains numbers nn and mm ( 1<=n,m<=501<=n,m<=50 ), nn — amount of lines, and mm — amount of columns on Bob's sheet. The following nn lines contain mm characters each. Character «.» stands for a non-shaded square on the sheet, and «*» — for a shaded square. It is guaranteed that Bob has shaded at least one square.
输出格式
Output the required rectangle of the minimum cost. Study the output data in the sample tests to understand the output format better.
题意翻译
题目描述
给定一 N \times MN×M 规模的矩阵，输出最小的包含所有 * 的矩阵。
输入格式
一行两个整数， NN 和 MM 。
然后一个 N \times MN×M 大小的矩阵。
输出格式
输出最小的包含所有 * 的矩阵。
数据范围
1 \leq N,M \leq 501≤N,M≤50。
输入输出样例
输入 #1
输出 #1
输入 #2
输出 #2
AC记录
AC__CODE
#include <iostream>
#define MAXN 51
using namespace std;
int n, m;
char a[MAXN][MAXN];
int u=0, d=0, l=0, r=0;
int main(){
cin >> n >> m;
for(int i=1; i<=n; i++)
for(int j=1; j<=m; j++)
cin >> a[i][j];
for(int i=1; i<=n; i++)
for(int j=1; j<=m; j++){
if(a[i][j] == '*'){
u = i;
break;
}
}
for(int i=n; i>=1; i--)
for(int j=1; j<=m; j++){
if(a[i][j] == '*'){
d = i;
break;
}
}
for(int j=1; j<=m; j++)
for(int i=1; i<=n; i++){
if(a[i][j] == '*'){
l = j;
break;
}
}
for(int j=m; j>=1; j--)
for(int i=1; i<=n; i++){
if(a[i][j] == '*'){
r = j;
break;
}
}
//cout << u << " " << d <<" "<< l << " "<< r << endl;
swap(u, d);
swap(l, r);
//cout << u << " " << d <<" "<< l << " "<< r << endl;
for(int i=u; i<=d; i++){
for(int j=l; j<=r; j++)
cout << a[i][j];
cout << endl;
}
return 0;
}