#644 (Div. 3)F. Spy-string（暴力枚举）

题目描述

You are given n strings a1,a2,…,an: all of them have the same length m. The strings consist of lowercase English letters.
Find any string s of length m such that each of the given n strings differs from s in at most one position. Formally, for each given string ai, there is no more than one position j such that ai[j]≠s[j].
Note that the desired string s may be equal to one of the given strings ai, or it may differ from all the given strings.
For example, if you have the strings abac and zbab, then the answer to the problem might be the string abab, which differs from the first only by the last character, and from the second only by the first.

Input

The first line contains an integer t (1≤t≤100) — the number of test cases. Then t test cases follow.
Each test case starts with a line containing two positive integers n (1≤n≤10) and m (1≤m≤10) — the number of strings and their length.
Then follow n strings ai, one per line. Each of them has length m and consists of lowercase English letters.

Output

Print t answers to the test cases. Each answer (if it exists) is a string of length m consisting of lowercase English letters. If there are several answers, print any of them. If the answer does not exist, print “-1” (“minus one”, without quotes).

Example

input
5
2 4
abac
zbab
2 4
aaaa
bbbb
3 3
baa
aaa
aab
2 2
ab
bb
3 1
a
b
c
output
abab
-1
aaa
ab
z

Note

The first test case was explained in the statement.
In the second test case, the answer does not exist.

题目大意

给你n个字符串，每个字符串长度为m，要求是否存在一个字符串使得该字符串与每个字符串的不同的字符个数小于等于1个。

题目分析

这道题就是暴力枚举每一种情况，然后验证是否合法即可。
其实就是考察写代码的能力。（比赛的时候我看出来是要暴力做了，但就是写不出来。。。）

代码如下

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <stack>
#include <map>
#include <unordered_map>
#include <queue>
#include <vector>
#include <set>
#include <algorithm>
#include <iomanip>
#define LL long long
using namespace std;
int const N=105;
int n,m;
string ans,s[15];
bool check()     //检查该情况是否合法
{for(int i=2;i<=m;i++)   //枚举其它m-1个字符串{int k=0;for(int j=0;j<n;j++){if(ans[j]!=s[i][j]) k++;   //统计不相同的字母数量if(k>=2) return false;     //大于2则不合法}}return true;    //否则合法
}
int main()
{int t;cin>>t;while(t--){cin>>m>>n;for(int i=1;i<=m;i++)cin>>s[i];bool st=false;for(int i=0;i<n;i++) //枚举每一位{ans=s[1];       //重置ans，保证每次只修改一个位置for(char j='a';j<='z';j++)   //枚举每一个字母{ans[i]=j;if(check()) {st=true; break;}  //检查是否合法，合法即可输出}if(st) break;}if(st) cout<<ans<<endl;   //如果有合法情况则输出else puts("-1");          //没有输出-1}return 0;
}

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#644 (Div. 3)F. Spy-string（暴力枚举）

题目描述

题目大意

题目分析

代码如下

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