hdu 4006 The kth great number，set，priority_queue

最新推荐文章于 2020-03-12 13:52:21 发布

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最新推荐文章于 2020-03-12 13:52:21 发布

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分类专栏： ACM-基础解题策略

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小明和小宝玩一个简单的数字游戏。小明可以选择写下数字或询问第k大的数字。由于写下的数字太多，小宝感到困惑。通过使用集合或优先队列维护长度为k的有序序列，我们可以帮助小宝找到所需的答案。

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The kth great number

Xiao Ming and Xiao Bao are playing a simple Numbers game. In a round Xiao Ming can choose to write down a number, or ask Xiao Bao what the kth great number is. Because the number written by Xiao Ming is too much, Xiao Bao is feeling giddy. Now, try to help Xiao Bao.
Input
There are several test cases. For each test case, the first line of input contains two positive integer n, k. Then n lines follow. If Xiao Ming choose to write down a number, there will be an " I" followed by a number that Xiao Ming will write down. If Xiao Ming choose to ask Xiao Bao, there will be a "Q", then you need to output the kth great number.
Output

The output consists of one integer representing the largest number of islands that all lie on one line.

Sample Input
8 3
I 1
I 2
I 3
Q
I 5
Q
I 4
Q
Sample Output
1
2
3

Hint
Xiao Ming won't ask Xiao Bao the kth great number when the number of the written number is smaller than k. (1=<k<=n<=1000000).

用set或者priority_queue维护长度为k从小到大有序的序列，每次查询的时候首元素即为所求。

#include <cstdio>
#include <set>
#include <map>
#include <vector>
#include <iostream>
using namespace std;

multiset<int> ms;

int main()
{
    int n, k, x;
    char ch[10];
    while(~scanf("%d%d", &n, &k)) {
        ms.clear();
        while(n--) {
            scanf("%s", ch);
            if(ch[0]=='I') {
                scanf("%d",&x);
                ms.insert(x);
                if(ms.size()>k) {
                    ms.erase(ms.begin());
                }
            } else {
                printf("%d\n", *ms.begin());
            }
        }
    }
    return 0;
}

#include <cstdio>
#include <set>
#include <map>
#include <vector>
#include <queue>
#include <iostream>
using namespace std;

priority_queue<int,vector<int>, greater<int> > pq;

int main()
{
    int n, k, x;
    char ch[10];
    while(~scanf("%d%d", &n, &k)) {
        while(!pq.empty()) pq.pop();
        while(n--) {
            scanf("%s", ch);
            if(ch[0]=='I') {
                scanf("%d",&x);
                pq.push(x);
                if(pq.size()>k) {
                    pq.pop();
                }
            } else {
                printf("%d\n", pq.top());
            }
        }
    }
    return 0;
}