Bombing

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others) Total Submission(s): 3492    Accepted Submission(s): 1323

Problem Description
It’s a cruel war which killed millions of people and ruined series of cities. In order to stop it, let’s bomb the opponent’s base. It seems not to be a hard work in circumstances of street battles, however, you’ll be encountered a much more difficult instance: recounting exploits of the military. In the bombing action, the commander will dispatch a group of bombers with weapons having the huge destructive power to destroy all the targets in a line. Thanks to the outstanding work of our spy, the positions of all opponents’ bases had been detected and marked on the map, consequently, the bombing plan will be sent to you. Specifically, the map is expressed as a 2D-plane with some positions of enemy’s bases marked on. The bombers are dispatched orderly and each of them will bomb a vertical or horizontal line on the map. Then your commanded wants you to report that how many bases will be destroyed by each bomber. Notice that a ruined base will not be taken into account when calculating the exploits of later bombers.
 
Input
Multiple test cases and each test cases starts with two non-negative integer N (N<=100,000) and M (M<=100,000) denoting the number of target bases and the number of scheduled bombers respectively. In the following N line, there is a pair of integers x and y separated by single space indicating the coordinate of position of each opponent’s base. The following M lines describe the bombers, each of them contains two integers c and d where c is 0 or 1 and d is an integer with absolute value no more than 109, if c = 0, then this bomber will bomb the line x = d, otherwise y = d. The input will end when N = M = 0 and the number of test cases is no more than 50.
 
Output
For each test case, output M lines, the ith line contains a single integer denoting the number of bases that were destroyed by the corresponding bomber in the input. Output a blank line after each test case.
 
Sample Input
3 2
1 2
1 3
2 3
0 1
1 3
0 0
 
Sample Output
2
1
 
Source
 
 
题意:输入 n,m
接着 n个点的 横纵坐标
m个询问(a,b) a为0代表 输出 横坐标为b的点的个数 并删除扫描过的点
                        a为1代表 输出 纵坐标为b的点的个数  并删除 扫描过的点
解法:map标记 set离散  有重点 用multiset
        STL大法好 涨姿势
 
注意 格式 PE  GG;
#include<bits/stdc++.h>
using namespace std;
int n,m;
int a,b;
map<int,multiset<int> >mp1;
map<int,multiset<int> >mp2;
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
if(n==0&&m==0)
break;
mp1.clear();
mp2.clear();
for(int i=1; i<=n; i++)
{
scanf("%d %d",&a,&b);
mp1[a].insert(b);
mp2[b].insert(a);
}
for(int i=1; i<=m; i++)
{
scanf("%d%d",&a,&b);
if(a==0)
{
printf("%d\n",mp1[b].size());
for(multiset<int>::iterator it=mp1[b].begin(); it!=mp1[b].end(); it++)
mp2[*it].erase(b);
mp1[b].clear();
}
else
{ printf("%d\n",mp2[b].size());
for(multiset<int>::iterator it=mp2[b].begin(); it!=mp2[b].end(); it++)
mp1[*it].erase(b);
mp2[b].clear();
}
}
printf("\n");
} return 0;
}

HDU4022 Bombing STL的更多相关文章

  1. HDU 4022 Bombing STL 模拟题

    人工模拟.. #include<stdio.h> #include<iostream> #include<algorithm> #include<vector ...

  2. HDU 4022 Bombing(stl,map,multiset,iterater遍历)

    题目 参考了     1     2 #define _CRT_SECURE_NO_WARNINGS //用的是STL中的map 和 multiset 来做的,代码写起来比较简洁,也比较好容易理解. ...

  3. [HDOJ4022]Bombing(离散化+stl)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4022 一个图上有n个点,之后m个操作,每次操作一行或者一列.使得这一行或者这一列的点全部消除.每次操作 ...

  4. 详细解说 STL 排序(Sort)

    0 前言: STL,为什么你必须掌握 对于程序员来说,数据结构是必修的一门课.从查找到排序,从链表到二叉树,几乎所有的算法和原理都需要理解,理解不了也要死记硬背下来.幸运的是这些理论都已经比较成熟,算 ...

  5. STL标准模板库(简介)

    标准模板库(STL,Standard Template Library)是C++标准库的重要组成部分,包含了诸多在计算机科学领域里所常见的基本数据结构和基本算法,为广大C++程序员提供了一个可扩展的应 ...

  6. STL的std::find和std::find_if

    std::find是用来查找容器元素算法,但是它只能查找容器元素为基本数据类型,如果想要查找类类型,应该使用find_if. 小例子: #include "stdafx.h" #i ...

  7. STL: unordered_map 自定义键值使用

    使用Windows下 RECT 类型做unordered_map 键值 1. Hash 函数 计算自定义类型的hash值. struct hash_RECT { size_t operator()(c ...

  8. C++ STL简述

    前言 最近要找工作,免不得要有一番笔试,今年好像突然就都流行在线笔试了,真是搞的我一塌糊涂.有的公司呢,不支持Python,Java我也不会,C有些数据结构又有些复杂,所以是时候把STL再看一遍了-不 ...

  9. codevs 1285 二叉查找树STL基本用法

    C++STL库的set就是一个二叉查找树,并且支持结构体. 在写结构体式的二叉查找树时,需要在结构体里面定义操作符 < ,因为需要比较. set经常会用到迭代器,这里说明一下迭代器:可以类似的把 ...

随机推荐

  1. [Clr via C#读书笔记]Cp10属性

    Cp10属性 属性的本质就是方法,只是看起来像字段罢了: 无参属性 就是一般属性: 字段一般要private,然后通过设置访问方法-访问器来访问:属性是方法语法变种:getset不一定要访问支持字段: ...

  2. jstat命令

    jstat命令使用 jstat命令可以查看堆内存各部分的使用量,以及加载类的数量.命令的格式如下: jstat [-命令选项] [vmid] [间隔时间/毫秒] [查询次数] 注意:使用的jdk版本是 ...

  3. 【zabbix 监控】第二章 安装测试被监控主机

    客户端安装测试 一.准备两台被监控主机,分别做如下操作: web129:192.168.19.129 web130:192.168.19.130 [root@web129 ~]#yum -y inst ...

  4. Windows10系统tensorflow-gpu安装

    准备工作 安装前请确保自己的显卡支持gpu加速,支持加速的gpu型号可在下面的链接中查询. https://www.geforce.com/hardware/technology/cuda/suppo ...

  5. Dev c++ 调试步骤

    不能调试的时候,修改下列地方: 1.在“工具”->编译选项->”Add following commands when calling complier”下面的编辑框里写入:-g3 2.在 ...

  6. leetcode个人题解——#20 Valid Parentheses

    class Solution { public: bool isValid(string s) { stack<char> brackts; ; i < s.size(); i++) ...

  7. Rescue(BFS时间最短 另开数组或优先队列)

    Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M ...

  8. Word Ladder Problem (DFS + BFS)

    Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest t ...

  9. Python 字符串与基本语句

    Python特点 python中没有变量的声明 语句结束后没有分号 严格要求缩进 支持很长很长的大数运算(直接在Idle中输入即可) 用"#"来注释 BIF:Bulit-in fu ...

  10. 软工网络15团队作业4——Alpha阶段敏捷冲刺-4

    一.当天站立式会议照片: 二.项目进展 昨天已完成的工作: 完成程序副界面的设计与信息的输入统计 明天计划完成的工作: 日期等细致信息的处理 工作中遇到的困难: 对微信小程序开发的代码构成有了一些了解 ...