来源poj1442 Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two…

优先队列..刚开始用蠢办法,经过一个vector容器中转,这么一来一回这么多趟,肯定超时啊. 超时代码如下: #include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <cstdlib> #include <string> #include <vector> #i…

题目地址:POJ 1442 这题是用了两个优先队列,当中一个是较大优先.还有一个是较小优先. 让较大优先的队列保持k个.每次输出较大优先队列的队头. 每次取出一个数之后,都要先进行推断,假设这个数比較大优先的队列的队头要小,就让它增加这个队列.队列头移到较小优先的队列中.然后当较大优先的数不足k个的时候,就让较小优先的队列的队头移到较大优先的队头中. 代码例如以下. #include <iostream> #include <cstdio> #include <string&…

题目链接:http://poj.org/problem?id=1442 思路分析: <1>维护一个最小堆与最大堆,最大堆中存储最小的K个数,其余存储在最小堆中; <2>使用Treap构造名次树,查询第K大数即可; 代码如下(堆的用法): #include<iostream> #include<queue> using namespace std; struct cmp1 { bool operator() ( const int a, const int b…

Description Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are t…

Black Box Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 7099   Accepted: 2888 Description Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empt…

Description Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are t…

Black Box Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 8637   Accepted: 3542 Description Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empt…

给n个数,依次按顺序插入,第二行m个数,a[i]=b表示在第b次插入后输出第i小的数 *解法:写两个优先队列,q1里由大到小排,q2由小到大排,保持q2中有i-1个元素,那么第i小的元素就是q2的top 而且注意到每次输出第i小的数i都是递增的 #include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <queue> #include…

题意:你有\(n\)个礼物,礼物有自己的种类,你想将它们按种类打包送人,但是打包的礼物数量必须不同(数量,与种类无关),同时,有些礼物你想自己留着,\(0\)表示你不想送人,问你在送出的礼物数量最大的同时,尽可能的使自己喜欢的留下来,输出能送出的最大礼物数,以及这些礼物中自己不喜欢的数目. 题解:首先,我们肯定要让送出的礼物数最大,同时喜欢的最小,也就是送的礼物中,尽量让它的\(f\)是\(1\).这题考虑贪心,我们可以先对礼物数量进行排序,礼物数量相同让\(1\)多的排在前面,全部丢进优先队列…