PAT L2-012. 关于堆的判断【数据结构】

题目链接

https://www.patest.cn/contests/gplt/L2-012

思路

题目本身不难,就是字符串处理有点繁琐。
但是有个巨坑!就是你必须得边push边造堆,不能一次性读完再造堆,两者造出来的顺序是不一样的!为此改了十多遍(累觉不爱)
这里用了STL的make_heap,自己手写也可以,不怎么长。

AC代码

#include <iostream>
#include <queue>
#include <vector>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <functional>
using namespace std;

vector<int>heap;
int find(int a)
{
    return distance(heap.begin(),find(heap.begin(),heap.end(),a));
}
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    while (n--)
    {
        int t;
        scanf("%d",&t);
        heap.push_back(t);
        make_heap(heap.begin(),greater<int>());
    }
    getchar();
    while (m--)
    {
        string q;
        getline(cin,q);
        if (q[q.length() - 1] == 't')
        {
            int a;
            stringstream ss(q);
            ss >> a;
            if (heap[0] == a) printf("T\n");
            else printf("F\n");
        }
        else if (q[q.length() - 1] == 's')
        {
            int a,b;
            string temp;
            stringstream ss(q);
            ss >> a >> temp >> b;
            int pos_a = find(a);
            int pos_b = find(b);
            pos_a++; pos_b++;
            if (pos_a / 2 == pos_b / 2)
            {
                printf("T\n");
            }
            else
            {
                printf("F\n");
            }
        }
        else
        {
            stringstream ss(q);
            int a,b;
            string temp;
            ss >> a >> temp >> temp;
            if (temp[0] == 't')
            {
                ss >> temp >> temp >> b;
                int pos_a = find(a);
                int pos_b = find(b);
                if (pos_b == pos_a * 2 + 1 || pos_b == pos_a * 2 + 2)
                {
                    printf("T\n");
                }
                else
                {
                    printf("F\n");
                }
            }
            else if (temp[0] == 'a')
            {
                ss >> temp >> temp >> b;
                int pos_a = find(a);
                int pos_b = find(b);
                if (pos_a == pos_b * 2 + 1 || pos_a == pos_b * 2 + 2)
                {
                    printf("T\n");
                }
                else
                {
                    printf("F\n");
                }
            }
        }
    }
    return 0;
}

相关文章

【啊哈!算法】算法3:最常用的排序——快速排序       ...
匿名组 这里可能用到几个不同的分组构造。通过括号内围绕的正...
选择排序:从数组的起始位置处开始,把第一个元素与数组中其...
public struct Pqitem { public int priority; ...
在编写正则表达式的时候,经常会向要向正则表达式添加数量型...
来自:http://blog.csdn.net/morewindows/article/details/6...