Codeforces 390A( 模拟题)
| Time Limit: 1000MS | Memory Limit: 262144KB | 64bit IO Format: %I64d & %I64u |
Description
Inna loves sleeping very much, so she needs n alarm clocks in total to wake up. Let's suppose that Inna's room is a 100 × 100 square with the lower left corner at point (0, 0) and with the upper right corner at point (100, 100). Then the alarm clocks are points with integer coordinates in this square.
The morning has come. All n alarm clocks in Inna's room are ringing, so Inna wants to turn them off. For that Inna has come up with an amusing game:
- First Inna chooses a type of segments that she will use throughout the game. The segments can be either vertical or horizontal.
- Then Inna makes multiple moves. In a single move, Inna can paint a segment of any length on the plane, she chooses its type at the beginning of the game (either vertical or horizontal), then all alarm clocks that are on this segment switch off. The game ends when all the alarm clocks are switched off.
Inna is very sleepy, so she wants to get through the alarm clocks as soon as possible. Help her, find the minimum number of moves in the game that she needs to turn off all the alarm clocks!
Input
The first line of the input contains integer n(1 ≤ n ≤ 105) — the number of the alarm clocks. The next n lines describe the clocks: the i-th line contains two integers xi, yi — the coordinates of the i-th alarm clock (0 ≤ xi, yi ≤ 100).
Note that a single point in the room can contain any number of alarm clocks and the alarm clocks can lie on the sides of the square that represents the room.
Output
In a single line print a single integer — the minimum number of segments Inna will have to draw if she acts optimally.
Sample Input
4
0 0
0 1
0 2
1 0
2
4
0 0
0 1
1 0
1 1
2
4
1 1
1 2
2 3
3 3
3
Hint
In the first sample, Inna first chooses type "vertical segments", and then she makes segments with ends at : (0, 0), (0, 2); and, for example,(1, 0), (1, 1). If she paints horizontal segments, she will need at least 3 segments.
In the third sample it is important to note that Inna doesn't have the right to change the type of the segments during the game. That's why she will need 3 horizontal or 3 vertical segments to end the game.
Source
#include <iostream>
#include <stdlib.h>
#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <math.h>
using namespace std;
typedef long long ll;
int main()
{
int i,j,m,n,sum1,sum2,x,y,a[],b[];
while (cin>>n)
{
sum1=;
sum2=;
memset(a,,sizeof(a));
memset(b,,sizeof(b));
for (i=;i<n;i++)
{
cin>>x>>y;
a[x]=;
b[y]=;
}
for (i=;i<;i++)
if (a[i])
sum1++;
for (i=;i<;i++)
if (b[i])
sum2++;
if (sum1<sum2)
cout<<sum1<<endl;
else
cout<<sum2<<endl;
}
return ;
}
Codeforces 390A( 模拟题)的更多相关文章
- CodeForces - 427B (模拟题)
Prison Transfer Time Limit: 1000MS Memory Limit: 262144KB 64bit IO Format: %I64d & %I64u Sub ...
- CodeForces - 404B(模拟题)
Marathon Time Limit: 1000MS Memory Limit: 262144KB 64bit IO Format: %I64d & %I64u Submit Sta ...
- CodeForces - 404A(模拟题)
Valera and X Time Limit: 1000MS Memory Limit: 262144KB 64bit IO Format: %I64d & %I64u Submit ...
- CodeForces 1B 模拟题。
H - 8 Time Limit:10000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Submit Statu ...
- Educational Codeforces Round 2 A. Extract Numbers 模拟题
A. Extract Numbers Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/600/pr ...
- Codeforces 767B. The Queue 模拟题
B. The Queue time limit per test:1 second memory limit per test:256 megabytes input:standard input o ...
- Codeforces Beta Round #7 B. Memory Manager 模拟题
B. Memory Manager 题目连接: http://www.codeforces.com/contest/7/problem/B Description There is little ti ...
- Codeforces Beta Round #5 B. Center Alignment 模拟题
B. Center Alignment 题目连接: http://www.codeforces.com/contest/5/problem/B Description Almost every tex ...
- Codeforces Beta Round #3 C. Tic-tac-toe 模拟题
C. Tic-tac-toe 题目连接: http://www.codeforces.com/contest/3/problem/C Description Certainly, everyone i ...
随机推荐
- 【BZOJ-2733】永无乡 Splay+启发式合并
2733: [HNOI2012]永无乡 Time Limit: 10 Sec Memory Limit: 128 MBSubmit: 2048 Solved: 1078[Submit][Statu ...
- BZOJ-1202 狡猾的商人 并查集+前缀和
我记得这个题,上次之前做的时候没改完,撂下了,今天突然想改发现,woc肿么A 了= =看来是我记错了.. 1202: [HNOI2005]狡猾的商人 Time Limit: 10 Sec Memory ...
- 宿主机( win 7 系统) ping 虚拟机VMware( cent os 6.6 ) 出现“请求超时”或者“无法访问目标主机”的解决方法
首先虚拟机的网络连接设置为"Host-only": 然后在 cmd 窗口中查看 VMnet1 的 ip 地址,这里是 192.168.254.1 接下来在 Linux 中设置网卡地 ...
- Linux Kernel中获取当前目录方法(undone)
目录 . 引言 . 基于进程内存镜像信息struct mm_struct获取struct path调用d_path()获取当前进程的"绝对路径" . 基于文件描述符(fd).tas ...
- maven运行javaWeb项目
首先从svn下载下来的maven项目,需要点击项目,然后import--->Existing Maven Projects->全选之后点next就转换成功了,然后 run as--> ...
- 最新版本的DBCP数据源配置
弄了我一下午,把该加的包都加进去了还是没用,后来把DBCP的包打开来看看才发现路径不对.配置如下: <!-- 使用dbcp配置数据源 --> <bean id="dataS ...
- android 常见死机问题--log分析
http://blog.csdn.net/fangchongbory/article/details/7645815 android 常见死机问题--log分析============ ...
- HD1561The more, The Better(树形DP+有依赖背包)
The more, The Better Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Oth ...
- latin1
Latin1是ISO-8859-1的别名,有些环境下写作Latin-1.ISO-8859-1编码是单字节编码,向下兼容ASCII,其编码范围是0x00-0xFF,0x00-0x7F之间完全和ASCII ...
- Visual Studio Online Integrations-Testing
原文:http://www.visualstudio.com/zh-cn/explore/vso-integrations-directory-vs