1. 题目链接

http://poj.org/problem?id=1442

2. 题目解读

可以利用大小堆，大堆长度从1开始，每次+1
大堆元素都比小堆的小，那么大堆顶的元素就是第k小的元素

3. 代码

3.1 Runtime Error 代码

本地运行示例，结果一致，poj提交RE，还没解决

/*** @description: 用大小堆求解* @author: michael ming* @date: 2019/5/31 23:21* @modified by: */
#include <iostream>
#include <algorithm>
#include <vector>
#include <functional>
using namespace std;
int arr[30005], total[30005];
int main()
{int arrlen, k, arrindex=1, maxheapsize=0, insertnum , minheapsize;cin >> arrlen >> k;for(int i = 1; i <= arrlen; ++i)cin >> arr[i];for(int i = 1; i <= k; ++i)cin >> total[i];vector<int> maxheap, minheap;for(int i = 1; i <= k; ++i){maxheapsize++;minheapsize = total[i] - maxheapsize;insertnum = total[i] - total[i-1];if(insertnum == 0 && !minheap.empty()){maxheap.push_back(minheap[0]);push_heap(maxheap.begin(), maxheap.end());//默认采用 < , 大堆pop_heap(minheap.begin(), minheap.end(), greater<int>());minheap.pop_back();}while(insertnum--){if (maxheap.empty()){maxheap.push_back(arr[arrindex]);}else{//----选择插入哪个堆-----if (arr[arrindex] <= maxheap[0]){if(maxheap.size() >= maxheapsize){minheap.push_back(maxheap[0]);//大堆顶进入小堆push_heap(minheap.begin(), minheap.end(), greater<int>());pop_heap(maxheap.begin(), maxheap.end());//堆顶到末尾了maxheap.pop_back();//删除到末尾的"堆顶"}maxheap.push_back(arr[arrindex]);push_heap(maxheap.begin(), maxheap.end());//默认采用 < , 大堆}else if (arr[arrindex] > maxheap[0]){if(minheap.size() >= minheapsize){maxheap.push_back(minheap[0]);push_heap(maxheap.begin(), maxheap.end());//默认采用 < , 大堆pop_heap(minheap.begin(), minheap.end(), greater<int>());minheap.pop_back();}minheap.push_back(arr[arrindex]);push_heap(minheap.begin(), minheap.end(), greater<int>());//小堆，采用 >}}arrindex++;}cout << maxheap[0] << endl;}return 0;
}