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 
      
       #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\n")using namespace std;const int maxn = 150000 + 5;vector
       
        
          > l, r;bool fl[maxn], fr[maxn];char ch[maxn];int n;int main(){    cin  >> n;    scanf("%s", ch);    vector
         
          
            > wenhaol;    for (int i = 0; ch[i] != '\0'; 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
           
            
              > wenhaor; for (int i = 0; ch[i] != '\0'; 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
             
              
                > ans; int lb = 0, rb = 0; while (lb < l.size() && rb < r.size()){ //printf("lb = %d rb = %d\n", 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\n", ans.size()); for (int i = 0; i < ans.size(); i++){ printf("%d %d\n", ans[i].fi + 1, ans[i].se + 1); } return 0;}