CodeForces 584C - Marina and Vasya(构造)

最新推荐文章于 2019-09-02 20:59:01 发布

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本文介绍了一道编程题的解决思路及代码实现。题目要求构造一个特殊字符串，使其与两个给定字符串在指定数量的位置上存在差异。文章详细解析了解题过程，并提供了一段完整的AC代码。

C. Marina and Vasya
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Marina loves strings of the same length and Vasya loves when there is a third string, different from them in exactly t characters. Help Vasya find at least one such string.

More formally, you are given two strings s1, s2 of length n and number t. Let’s denote as f(a, b) the number of characters in which strings a and b are different. Then your task will be to find any string s3 of length n, such that f(s1, s3) = f(s2, s3) = t. If there is no such string, print - 1.

Input
The first line contains two integers n and t (1 ≤ n ≤ 105, 0 ≤ t ≤ n).

The second line contains string s1 of length n, consisting of lowercase English letters.

The third line contain string s2 of length n, consisting of lowercase English letters.

Output
Print a string of length n, differing from string s1 and from s2 in exactly t characters. Your string should consist only from lowercase English letters. If such string doesn’t exist, print -1.

Examples
input
3 2
abc
xyc
output
ayd
input
1 0
c
b
output
-1

题意:
对于长度为n的两个字符串,要构造一个新的字符串,使得他与任意的字符串都有对应位置的t个不同字符.

解题思路:
对于这个字符串,必然和原字符串有n-t个相同.
那么在找相同的时候,选择a和b相同的是最优解.
记录a和b相同的个数为sum.
那么对于a和新字符串,他们最多的相同个数为sum + (n-sum)/2 (因为也需要和b有(n-sum)/2相同).
所以最大相同个数如果小于n-t,必然是不成立的.

在优先填充完sum部分后,再分别对于a和b进行单独相同填充.
最后的剩余部分要使得与a和b都不相同.

AC代码:

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+5;
char a[maxn];
char b[maxn];
char c[maxn];
int main()
{
    int n,t;
    cin>>n>>t;
    cin>>a>>b;
    int sum = 0;
    for(int i = 0;a[i];i++) if(a[i] == b[i])    sum++;
    int m = n-t;
    if(sum + (n-sum)/2 < m)
    {
        printf("-1");
        return 0;
    }
    int k = min(sum,m);
    int cnt = 0;
    for(int i = 0,cnt = 1;a[i]&&cnt <= k;i++)   if(a[i] == b[i])    c[i] = a[i],cnt++;
    int s = m-k;
    for(int i = 0,cnt = 1;a[i] && cnt <= s;i++) if((a[i] != b[i]) && !c[i]) c[i] = a[i],cnt++;
    for(int i = 0,cnt = 1;a[i] && cnt <= s;i++)  if((a[i] != b[i]) && !c[i]) c[i] = b[i],cnt++;
    for(int i = 0;a[i];i++)
    {
        if(!c[i])
        {
            for(char j = 'a';j <= 'z';j++)
                if(j != a[i] && j != b[i])  {c[i] = j;break;}
        }
    }
    cout<<c;
    return 0;
}