链接:https://ac.nowcoder.com/acm/contest/908/F
题意:
AFei has many cards. Each card has a number written on it. Now he wants to takes some out of his card and puts them in a box. And he wants to know whether the card with the number x was in the box. So he has the following two operations:
- 0 x (It means to put a card with the number x in the box.)
- 1 x (It means to query if there is a card with the number x in the box.
思路:
map超时。。离散化,离线处理。
代码:
#include <bits/stdc++.h> using namespace std; typedef long long LL; const int MAXN = 1e6 + 10; const int MOD = 1e9 + 7; int n, m, k, t; struct Node { int v, pos; bool operator < (const Node& that) const { return this->v < that.v; } }node[MAXN]; int a[MAXN]; int vis[MAXN]; int op[MAXN]; int main() { scanf("%d", &n); for (int i = 1;i <= n;i++) { scanf("%d%d", &op[i], &node[i].v); node[i].pos = i; } sort(node+1, node+1+n); int cnt = 1; a[node[1].pos] = cnt; for (int i = 2;i <= n;i++) { if (node[i].v == node[i-1].v) a[node[i].pos] = cnt; else a[node[i].pos] = ++cnt; } for (int i = 1;i <= n;i++) { if (op[i] == 0) vis[a[i]] = 1; else if (vis[a[i]] == 1) printf("yes "); else printf("no "); } return 0; }