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

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題目大意：

鞋子匹配，用一個小寫字母表示一種顏色。L[i]表示左腳的顏色，R[i]表示右腳的顏色，只有當L[i]和R[j]的顏色差不多了，才算匹配成功。但是，有一種特殊的顏色‘？’，該顏色可以和任意另一半鞋子匹配。

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思路：

取出‘？’，格外判斷就好了

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//看看會不會爆int!數(shù)組會不會少了一維！ //取物問題一定要小心先手勝利的條件 #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\n")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] != '\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<pair<int, int> > 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<pair<int, int> > 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; } View Code

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轉載于:https://www.cnblogs.com/heimao5027/p/10650300.html

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