Almost Rectangle_综合

题目

There is a square field of size n×n in which two cells are marked. These cells can be in the same row or column.

You are to mark two more cells so that they are the corners of a rectangle with sides parallel to the coordinate axes.

For example, if n=4 and a rectangular field looks like this (there are asterisks in the marked cells):

…?.
…
?…
…
Then you can mark two more cells as follows

?.?.
…
?.?.
…
If there are several possible solutions, then print any of them.

Input
The first line contains a single integer t (1≤t≤400). Then t test cases follow.

The first row of each test case contains a single integer n (2≤n≤400) — the number of rows and columns in the table.

The following n lines each contain n characters ‘.’ or ‘*’ denoting empty and marked cells, respectively.

It is guaranteed that the sums of n for all test cases do not exceed 400.

It is guaranteed that there are exactly two asterisks on the field. They can be in the same row/column.

It is guaranteed that the solution exists.

Output
For each test case, output n rows of n characters — a field with four asterisks marked corresponding to the statements. If there multiple correct answers, print any of them.

样例输入

Example
Input
6
4
….
…
…
…
2
.
.
2
.
.
3
.
…
…
5
…
……
…
.…
…
4
…
…
*…
*…

样例格式

Output
..
…
..
…
**
**
**
**
.
.
…
…
.…
…
.…
…
…
…
**…
**…

数据范围

不大于400

题意

每个矩阵里有两个星号，再添加两个星号，使四个星号组成一个矩阵

思路

1.给的两个星号不同行不同列的情况

2.给的两个星号同行不同列，在第一行和最后一行的情况

3.给的两个星号不同行同列，在第一列和最后一列的情况

坑点

1.给的两个星号同行不同列，在第一行和最后一行的情况

2.给的两个星号不同行同列，在第一列和最后一列的情况

代码

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int k;cin>>k;while(k--){
    int n;char num[405][405];cin>>n;int a,b,c,d;bool t=0;for(int i=0;i<n;i++){
    for(int j=0;j<n;j++){
    cin>>num[i][j];if(num[i][j]=='*'&&t==0){
    a=i;b=j;t=1;}if(num[i][j]=='*'&&t==1){
    c=i;d=j;}}}if(n==2){
    for(int i=0;i<n;i++){
    for(int j=0;j<n;j++){
    cout<<'*';}cout<<endl;}}else{
    if(a!=c&&b!=d){
    num[a][d]='*';num[c][b]='*';}else if(a==c&&b!=d&&a!=n-1){
    num[a+1][d]='*';num[c+1][b]='*';}else if(a==c&&b!=d&&a==n-1){
    num[a-1][d]='*';num[c-1][b]='*';}else if(a!=c&&b==d&&b!=n-1){
    num[a][d+1]='*';num[c][b+1]='*';}else if(a!=c&&b==d&&b==n-1){
    num[a][d-1]='*';num[c][b-1]='*';}for(int i=0;i<n;i++){
    for(int j=0;j<n;j++){
    cout<<num[i][j];}cout<<endl;}}}return 0;
}

总结

用scanf和printf输入输出会比cin,cout快