F.Cards with Numbers

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\n");
            else
                printf("no\n");
    }

    return 0;
}

原文地址：https://www.cnblogs.com/YDDDD/p/10960375.html