[HDOJ4006]The kth great number

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4006

 

提供插入查询两个操作，问这些数中第k大的数字是几。

可以用最小堆完成这个任务，把k看作是堆的容量，堆顶就是第k大的数。这样只需要保留前k大的数输出最小那个就可以了。

#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <climits>
#include <complex>
#include <fstream>
#include <cassert>
#include <cstdio>
#include <bitset>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <ctime>
#include <set>
#include <map>
#include <cmath>

using namespace std;

const int maxn = 100010;
const int INF = 2147483647;
int heap[maxn];
int pos;

void init() {
    pos = 0;
    memset(heap, 0, sizeof(heap));
    heap[0] = -INF;
}

void push(int x) {
    int i = ++pos;
    for(; heap[i>>1] > x; i>>=1) {
        heap[i] = heap[i>>1];
    }
    heap[i] = x;
}

void pop() {
    if(pos == 0) return;
    int child = 1;
    int i = 1;
    int last = heap[pos--];
    for(; i<<1 <= pos; i=child) {
        child = i<<1;
        if(child != pos && heap[child] > heap[child+1]) {
            ++child;
        }
        if(last > heap[child]) {
            heap[i] = heap[child];
        }
        else {
            break;
        }
    }
    heap[i] = last;
}

int n, k, x;
char cmd[2];

int main() {
    // freopen("in", "r", stdin);
    while(~scanf("%d %d", &n, &k)) {
        init();
        while(n--) {
            scanf("%s", cmd);
            if(cmd[0] == 'I') {
                scanf("%d", &x);
                if(pos < k) push(x);
                else if(heap[1] < x) {
                    pop();
                    push(x);
                }
            }
            else {
                printf("%d\n", heap[1]);
            }
        }
    }
    return 0;
}