poj 2623 Sequence Median 堆的灵活运用】的更多相关文章

I - Sequence Median Time Limit:1000MS     Memory Limit:1024KB     64bit IO Format:%I64d & %I64u Submit Status Description Given a sequence of N nonnegative integers. Let's define the median of such sequence. If N is odd the median is the element with…
Description Given a sequence of N nonnegative integers. Let's define the median of such sequence. If N is odd the median is the element with stands in the middle of the sequence after it is sorted. One may notice that in this case the median has posi…
题目链接:http://poj.org/problem?id=2442 题目大意:给出一个m*n的矩阵,从每一行中取出一个数相加.能得到n^m个不同的结果.要求输出当中前n项. 建立一个以n元数组为底层数组的堆,在这里,利用stl中的make_heap,pop_heap.push_heap等函数解决. 1.将第一组数据输入arr1数组.升序排序. 2.将接下来的数据输入到arr2数组中.而且heap[i]=arr1[0]+arr2[0...n-1].make_heap(heap,heap+n).…
1306. Sequence Median Time limit: 1.0 secondMemory limit: 1 MBLanguage limit: C, C++, Pascal Given a sequence of N nonnegative integers. Let's define the median of such sequence. If N is odd the median is the element with stands in the middle of the…
[POJ 3581]Sequence 标签: 后缀数组 题目链接 题意 给你一串序列\(A_i\),保证对于$ \forall i \in [2,n],都有A_1 >A_i$. 现在需要把这个序列分成三段,并且将这三段分别翻转,求如何翻转使整个序列字典序最小.(每一段不能为空) 题解 首先可以确定第一段的位置. 注意到,\(A_1\)是最大的,所以我们就只用考虑怎样找到一个前缀使其翻转后的字典序最小. (假如不是的话,就可能找到两个前缀翻转之后,一个为另一个的前缀,无法解决) 这等价于翻转之后找…
For this problem, you will write a program that reads in a sequence of 32-bit signed integers. After each odd-indexed value is read, output the median (middle value) of the elements received so far. Input The first line of input contains a single int…
题目链接:http://poj.org/problem?id=2442 Time Limit: 6000MS Memory Limit: 65536K Description Given m sequences, each contains n non-negative integer. Now we may select one number from each sequence to form a sequence with m integers. It's clear that we ma…
题目地址:id=2442">POJ 2442 真心没想到这题的思路. .原来是从第一行逐步向下加,每次都仅仅保存前n小的数.顺便练习了下堆.. 只是感觉堆的这样的使用方法用的不太多啊.. 又是手残. . 把j写成了i,于是就改啊改..改的跟题解上的差点儿一样了= = !. . 代码例如以下: #include <iostream> #include <cstdio> #include <string> #include <cstring> #…
题目描述 给定m个序列,每个序列包含n个非负整数.现在我们可以从每个序列中选择一个数字以形成一个具有m个整数的序列.显然,我们可以得到n ^ m种这种序列.然后,我们可以计算每个序列中的数字总和,并获得n ^ m个值.我们需要的是最小的n个和.你可以帮我们吗? 题目大意:给定M个长度为N的序列,从每个序列中任意取一个数求和,可以构成N的M次方个和,求其中最小的N个和. 输入格式 第一行是整数T,它显示测试用例的数量,然后是T个测试用例.每种情况的第一行都包含两个整数m,n(0 <m <= 10…
Description For this problem, you will write a program that reads in a sequence of 32-bit signed integers. After each odd-indexed value is read, output the median (middle value) of the elements received so far. Input The first line of input contains…