Codeforces Round #547 (Div. 3) D

http://codeforces.com/contest/1141/problem/D

题目大意：

鞋子匹配，用一个小写字母表示一种颜色。L[i]表示左脚的颜色，R[i]表示右脚的颜色，只有当L[i]和R[j]的颜色差不多了，才算匹配成功。但是，有一种特殊的颜色‘？’，该颜色可以和任意另一半鞋子匹配。

思路：

取出‘？’，格外判断就好了

//看看会不会爆int!数组会不会少了一维！
//取物问题一定要小心先手胜利的条件
#include <bits/stdc++.h>
#pragma comment(linker,"/STACK:102400000,102400000")
#define LL long long
#define ALL(a) a.begin(), a.end()
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define haha printf("haha
")

using namespace std;
const int maxn = 150000 + 5;
vector<pair<int, int> > l, r;
bool fl[maxn], fr[maxn];
char ch[maxn];
int n;


int main(){
    cin  >> n;
    scanf("%s", ch);
    vector<pair<int, int> > wenhaol;
    for (int i = 0; ch[i] != ''; i++){
        if (ch[i] >= 'a' && ch[i] <= 'z')
            l.pb(mk(ch[i], i));
        else if (ch[i] == '?')
            wenhaol.pb(mk(ch[i] + 80, i));
    }
    sort(l.begin(), l.end());


    scanf("%s", ch);
    vector<pair<int, int> > wenhaor;
    for (int i = 0; ch[i] != ''; i++){
        if (ch[i] >= 'a' && ch[i] <= 'z')
            r.pb(mk(ch[i], i));
        else if (ch[i] == '?')
            wenhaor.pb(mk(ch[i] + 80, i));
    }
    sort(r.begin(), r.end());

    vector<pair<int, int> > ans;
    int lb = 0, rb = 0;
    while (lb < l.size() && rb < r.size()){
        //printf("lb = %d rb = %d
", lb, rb);
        if (l[lb].fi == r[rb].fi){
            ans.pb(mk(l[lb].se, r[rb].se));
            fl[lb] = fr[rb] = 1;
            lb++, rb++;
            continue;
        }
        /*if (l[lb].fi == '?' + 80 || r[rb].fi == '?' + 80){
            ans.pb(mk(l[lb].se, r[rb].se));
            fl[lb] = fr[rb] = 1;
            lb++, rb++;
            continue;
        }*/
        if (l[lb].fi > r[rb].fi && l[lb].fi != '?' + 80){
            rb++;
            continue;
        }
        if (r[rb].fi > l[lb].fi && r[rb].fi != '?' + 80){
            lb++;
            continue;
        }
    }

    lb = 0, rb = 0;
    for (int i = 0; i < l.size(); i++){
        if (fl[i] != 1 && rb < wenhaor.size()){
            ans.pb(mk(l[i].se, wenhaor[rb].se));
            rb++;

        }
    }

    for (int i = 0; i < r.size(); i++){
        if (fr[i] != 1 && lb < wenhaol.size()){
            ans.pb(mk(wenhaol[lb].se, r[i].se));
            lb++;
        }
    }

    while (lb < wenhaol.size() && rb < wenhaor.size()){
        ans.pb(mk(wenhaol[lb].se, wenhaor[rb].se));
        lb++, rb++;
    }


    printf("%d
", ans.size());
    for (int i = 0; i < ans.size(); i++){
        printf("%d %d
", ans[i].fi + 1, ans[i].se + 1);
    }


    return 0;
}