codeforces 324# C. Marina and Vasya (贪心)

题目：http://codeforces.com/contest/584/problem/C

题意：给定两个字符串s1和s2，求字符串s3使得s3与s1对应位置字符不同的个数为t，并且s3与s2对应位置字符不同的个数也为t。

C. Marina and Vasya

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard 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.

Sample test(s)

input

3 2
abc
xyc

output

ayd

input

1 0
c
b

output

-1

分析：如果构造不出s3，就输出-1。由于找一个与另外两个字符的字符很好找，那么构造不出s3的情况是，s3和s1，s3和s2对应位置相同字符的个数不足。那么最大限度地利用一个位置的贡献就好了，怎么最大限度利用一个位置？当s1[i]==s2[i]的时候，把s3[i]赋成s1[i]。剩下的位置s1和s2交替贡献字符就行了。最后看满不满足条件。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;
const int maxn = 1e6;
char s1[maxn],s2[maxn],ans[maxn];
int main()
{
int n,t,i,j,lack;
cin>>n>>t;
cin>>s1>>s2;
lack=n-t;
int cnt1=0,cnt2=0;
memset(ans,'\0',sizeof(ans));
for(i=0;i<n;i++)
{
if(cnt1==lack)
break;
if(s1[i]==s2[i])
{
ans[i]=s1[i];
s1[i]=s2[i]='#';
cnt1++;
cnt2++;
}
}
// printf("%s %s %s\n",ans,s1,s2);
int fg=1;
for(i=0;i<n;i++) if(s1[i]!='#')
{
if(cnt1==lack && cnt2==lack)
break;
if(cnt1==cnt2)
{
ans[i]=s1[i];
cnt1++;
}
else
{
ans[i]=s2[i];
cnt2++;
}
s1[i]=s2[i]='#';
}
// printf("%s %s %s",ans,s1,s2);
for(i=0;i<n;i++) if(s1[i]!='#')
{
for(char ch='a';ch<='z';ch++)
{
if(ch!=s1[i] && ch!=s2[i])
{
ans[i]=ch;
break;
}
}
}
if(cnt1!=lack || cnt2!=lack)
{
printf("-1\n");
return 0;
}
printf("%s\n",ans);
return 0;
}