D. Colored Boots(div3 stl)

来源:https://codeforces.com/contest/1141/problem/D

There are nn left boots and nn right boots. Each boot has a color which is denoted as a lowercase Latin letter or a question mark ('?'). Thus, you are given two strings ll and rr, both of length nn. The character lili stands for the color of the ii-th left boot and the character riri stands for the color of the ii-th right boot.

A lowercase Latin letter denotes a specific color, but the question mark ('?') denotes an indefinite color. Two specific colors are compatibleif they are exactly the same. An indefinite color is compatible with any (specific or indefinite) color.

For example, the following pairs of colors are compatible: ('f', 'f'), ('?', 'z'), ('a', '?') and ('?', '?'). The following pairs of colors are notcompatible: ('f', 'g') and ('a', 'z').

Compute the maximum number of pairs of boots such that there is one left and one right boot in a pair and their colors are compatible.

Print the maximum number of such pairs and the pairs themselves. A boot can be part of at most one pair.

Input

The first line contains nn (1≤n≤1500001≤n≤150000), denoting the number of boots for each leg (i.e. the number of left boots and the number of right boots).

The second line contains the string ll of length nn. It contains only lowercase Latin letters or question marks. The ii-th character stands for the color of the ii-th left boot.

The third line contains the string rr of length nn. It contains only lowercase Latin letters or question marks. The ii-th character stands for the color of the ii-th right boot.

Output

Print kk — the maximum number of compatible left-right pairs of boots, i.e. pairs consisting of one left and one right boot which have compatible colors.

The following kk lines should contain pairs aj,bjaj,bj (1≤aj,bj≤n1≤aj,bj≤n). The jj-th of these lines should contain the index ajaj of the left boot in the jj-th pair and index bjbj of the right boot in the jj-th pair. All the numbers ajaj should be distinct (unique), all the numbers bjbj should be distinct (unique).

If there are many optimal answers, print any of them.

Examples

input

10
codeforces
dodivthree

output

5
7 8
4 9
2 2
9 10
3 1

input

7
abaca?b
zabbbcc

output

5
6 5
2 3
4 6
7 4
1 2

input

9
bambarbia
hellocode

output

0

input

10
code??????
??????test

output

10
6 2
1 6
7 3
3 5
4 8
9 7
5 1
2 4
10 9
8 10

题意: 给两个字符串a和b,每个字符表示一个靴子的颜色,两个相同的字母被认为是一对合法的靴子,如[f,f],[a,a],问号字符表示未知颜色的靴子,问号可以和任何颜色的靴子合法配对,问号也能和问号配对如,(a,?),(?,b),(?,?),问两个字符串最大能匹配多少对靴子,输出每对靴子的下标。

模拟就好了,用26个容器存储所有靴子的下标,然后直接遍历a和b串中是否存在某个字母,存在就塞入ans容器中。

注意最后对比一下问号容器中是否存在能和不成对颜色组成一对的情况。

#include<iostream>
#include<vector>
#include<stack>
using namespace std;
const int maxn=150005;
int n;
string a,b;
stack<int>st1[30],st2[30];
vector<pair<int,int> >ans;
int main()
{
    cin>>n>>a>>b;
    for(int i=0; i<n; i++)
    {
        if(a[i]=='?')st1[27].push(i+1);
        if(b[i]=='?')st2[27].push(i+1);
        if(a[i]>='a'&&a[i]<='z')st1[a[i]-'a'].push(i+1);
        if(b[i]>='a'&&b[i]<='z')st2[b[i]-'a'].push(i+1);
    }
    for(int i=0; i<26; i++)
    {
        while(!st1[i].empty()&&!st2[i].empty())//字母
        {
            ans.push_back(make_pair(st1[i].top(),st2[i].top()));
            st1[i].pop();
            st2[i].pop();
        }
        while(!st1[i].empty()&&!st2[27].empty())//未配对的字母和?
        {
            ans.push_back(make_pair(st1[i].top(),st2[27].top()));
            st1[i].pop();
            st2[27].pop();
        }
        while(!st2[i].empty()&&!st1[27].empty())//?和未配对的字母
        {
            ans.push_back(make_pair(st1[27].top(),st2[i].top()));
            st1[27].pop();
            st2[i].pop();
        }
    }
    while(!st1[27].empty()&&!st2[27].empty())//未配对的??
    {
        ans.push_back(make_pair(st1[27].top(),st2[27].top()));
        st1[27].pop();
        st2[27].pop();
    }
    cout<<ans.size()<<endl;
    while(ans.size())
    {
        cout<<ans.back().first<<" "<<ans.back().second<<endl;
        ans.pop_back();
    }
    return 0;
}

 

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03-26 13:04
已编辑
电子科技大学 算法工程师
xiaowl:你这个简历“条目上”都比较有深度性,但是实际上面试官又没法很好的评估你是怎么达到很多看上去很厉害的结果的。要避免一些看上去很厉害的包装,比如高效的内存复用策略的表达,如果仅是简单的一些内存共享机制,而且面试上也没有深挖的空间,就不要这样表达。比如,工程化模式本质上可能就是定义了一些abstract class,那也就没特别多值得讲的内容。建议简历上应该侧重那些你花了大量时间和精力解决、研究的问题,不要过分追求“丰富”,而是关注在技术深入度、问题解决能力的表现上。
没有实习经历,还有机会进...
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