标签:oid name max online ++ continue tin ras ace
[题目链接]
https://www.lydsy.com/JudgeOnline/problem.php?id=2028
[算法]
直接用std :: set维护即可
时间复杂度 : O(NlogN)
[代码]
#include<bits/stdc++.h> using namespace std; int n; struct segment { int l , r; friend bool operator < (segment a , segment b) { if (a.r == b.r) return a.l < b.l; else return a.r < b.r; } } ; template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); } template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); } template <typename T> inline void read(T &x) { T f = 1; x = 0; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == ‘-‘) f = -f; for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - ‘0‘; x *= f; } int main() { scanf("%d" , &n); set< segment > S; for (int i = 1; i <= n; i++) { char type[5]; scanf("%s" , &type); if (type[0] == ‘B‘) { printf("%d\n" , (int)S.size()); continue; } else { int ans = 0; int l , r; scanf("%d%d" , &l , &r); while (true) { set< segment > :: iterator it = S.lower_bound((segment){0 , l}); if (it == S.end()) break; segment tmp = (*it); if (tmp.l <= r) { S.erase(it); ++ans; } else break; } printf("%d\n" , ans); S.insert((segment){l , r}); } } return 0; }
标签:oid name max online ++ continue tin ras ace
原文地址:https://www.cnblogs.com/evenbao/p/9902314.html