C - To Be an Dream Architect

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Appoint description: 
System Crawler  (2014-11-05)

Description

The “dream architect” is the key role in a team of “dream extractors” who enter other’s dreams to steal secrets. A dream architect is responsible for crafting the virtual world that the team and the target will dream into. To avoid the target noticing the world is artificial, a dream architect must have powerful 3D imagination.

Cobb uses a simple 3D imagination game to test whether a candidate has the potential to be an dream architect. He lets the candidate imagine a cube consisting of n×n×n blocks in a 3D coordinate system as Figure 1. The block at bottom left front corner is marked (1, 1, 1) and the diagonally opposite block is marked (n, n, n). Then he tells the candidate that the blocks on a certain line are eliminated. The line is always parallel to an axis. After m such block eliminations, the candidate is asked to tell how many blocks are eliminated. Note that one block can only be eliminated once even if it is on multiple lines.

Here is a sample graph according to the first test case in the sample input: 

 

Input

The first line is the number of test cases. 
In each test case, the first line contains two integers n and m( 1 <= n <= 1000, 0 <= m <= 1000).,meaning that the cube is n x n x n and there are m eliminations.

Each of the following m lines represents an elimination in the following format: 
axis_1=a, axis_2=b 
where axis_i (i=1, 2) is ‘X’ or ‘Y’, or ‘Z’ and axis_1 is not equal to axis_2. a and b are 32-bit signed integers. 

 

Output

For each test case output the number of eliminated blocks.
 

Sample Input

2
3 2
Y=1,Z=3
X=3,Y=1
10 2
X=3,Y=3
Y=3,Z=3
 

Sample Output

5
19

感想:简单的容斥原理,但是统计得头晕眼花,输入也给人找麻烦,姿势还是不好

思路:每次消掉一行,比如消掉的是坐标为(x0,y0),平行z轴的某行,也就是不能有坐标为(x0,y0,*)的三维整数点满足题意,同理,对于(*,y0,z0),(x0,*,z0)也是,(*通配),所以容斥,得到总数

#include <cstdio>
#include <cstring>
using namespace std;
int n,m;
int a,b,c,d;
bool xy[1001][1001],xz[1001][1001],yz[1001][1001];
int xylen[1001],yxlen[1001],yzlen[1001],zxlen[1001],zylen[1001],xzlen[1001];
char buff[100];
int xzheap[1001][2];
int read1(){
int ans=0;
for(int i=2;i<=6;i++){
if(buff[i]>'9'||buff[i]<'0')break;
ans*=10;
ans+=buff[i]-'0';
}
return ans;
}
int read2(){
int ans=0;
int i=6;
while(buff[i]>'9'||buff[i]<'0'){i++;}
for(;;i++){
if(buff[i]>'9'||buff[i]<'0')break;
ans*=10;
ans+=buff[i]-'0';
}
return ans;
}
void inset(char ch1,char ch2,int num1,int num2){
if(ch1=='X'){
if(ch2=='Y'){
if(!xy[num1][num2]){
a++;
xy[num1][num2]=true;
xylen[num1]++;
yxlen[num2]++;
}
}
if(ch2=='Z'){
if(!xz[num1][num2]){
xzheap[b][0]=num1;
xzheap[b++][1]=num2;
xz[num1][num2]=true;
xzlen[num1]++;
zxlen[num2]++;
}
}
}
if(ch1=='Y'){
if(ch2=='X'){
if(!xy[num2][num1]){
a++;
xy[num2][num1]=true;
xylen[num2]++;
yxlen[num1]++;
}
}
if(ch2=='Z'){
if(!yz[num1][num2]){
c++;
yz[num1][num2]=true;
yzlen[num1]++;
zylen[num2]++;
}
}
}
if(ch1=='Z'){
if(ch2=='X'){
if(!xz[num2][num1]){
xzheap[b][0]=num2;
xzheap[b++][1]=num1;
xz[num2][num1]=true;
xzlen[num2]++;
zxlen[num1]++;
}
}
if(ch2=='Y'){
if(!yz[num2][num1]){
c++;
yz[num2][num1]=true;
yzlen[num2]++;
zylen[num1]++;
}
}
}
}
void cl2ear(){
a=b=c=d=0;
memset(xy,0,sizeof(xy));
memset(yz,0,sizeof(yz));
memset(xz,0,sizeof(xz));
memset(xylen,0,sizeof(xylen));
memset(yxlen,0,sizeof(xylen));
memset(zylen,0,sizeof(xylen));
memset(yzlen,0,sizeof(xylen));
memset(xzlen,0,sizeof(xylen));
memset(zxlen,0,sizeof(xylen));
}
int main(){
int T;
scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&m);
cl2ear();
for(int i=0;i<m;i++){
scanf("%s",buff);
bool fl=false;
char ch1,ch2;int num1=read1(),num2=read2();
for(int j=0;buff[j];j++){
if(buff[j]=='X'||buff[j]=='Y'||buff[j]=='Z'){
if(!fl){
ch1=buff[j];
fl=true;
}
else {
ch2=buff[j];
break;
}
}
}
inset(ch1,ch2,num1,num2);
}
for(int i=0;i<b;i++){
for(int j=1;j<=n;j++){
if(xy[xzheap[i][0]][j]&&yz[j][xzheap[i][1]])d++;
}
}
int ans=n*a+n*b+n*c+d;
int a=0;
for(int i=1;i<=n;i++){
a+=xylen[i]*xzlen[i]+yzlen[i]*yxlen[i]+zxlen[i]*zylen[i];
}
ans-=a;
printf("%d\n",ans);
}
return 0;
}

  

hdu 3682 10 杭州 现场 C To Be an Dream Architect 容斥 难度:0的更多相关文章

  1. hdu 3682 10 杭州 现场 C - To Be an Dream Architect 简单容斥 难度:1

    C - To Be an Dream Architect Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d &a ...

  2. hdu 4771 13 杭州 现场 B - Stealing Harry Potter's Precious 暴力bfs 难度:0

    Description Harry Potter has some precious. For example, his invisible robe, his wand and his owl. W ...

  3. hdu 3685 10 杭州 现场 F - Rotational Painting 重心 计算几何 难度:1

    F - Rotational Painting Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & % ...

  4. hdu 3687 10 杭州 现场 H - National Day Parade 水题 难度:0

    H - National Day Parade Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & % ...

  5. hdu 3695 10 福州 现场 F - Computer Virus on Planet Pandora 暴力 ac自动机 难度:1

    F - Computer Virus on Planet Pandora Time Limit:2000MS     Memory Limit:128000KB     64bit IO Format ...

  6. hdu 3697 10 福州 现场 H - Selecting courses 贪心 难度:0

    Description     A new Semester is coming and students are troubling for selecting courses. Students ...

  7. hdu 3699 10 福州 现场 J - A hard Aoshu Problem 暴力 难度:0

    Description Math Olympiad is called “Aoshu” in China. Aoshu is very popular in elementary schools. N ...

  8. hdu 3696 10 福州 现场 G - Farm Game DP+拓扑排序 or spfa+超级源 难度:0

    Description “Farm Game” is one of the most popular games in online community. In the community each ...

  9. hdu 3694 10 福州 现场 E - Fermat Point in Quadrangle 费马点 计算几何 难度:1

    In geometry the Fermat point of a triangle, also called Torricelli point, is a point such that the t ...

随机推荐

  1. JSP学习(第一课)

    JSP页面组成: 比如: 打开网页,右键查看源代码: 打开网页: 注意: <%!%>里面定义的属性是成员属性,相当于类的属性,方法相当于是全局的方法,相当于是类里面的方法.但是它是不可以进 ...

  2. 实现linux下的ls

    实现linux下的ls ls的使用 ls -a 列出文件下所有的文件,包括以"."开头的隐藏文件(linux下文件隐藏文件是以.开头的,如果存在..代表存在着父目录). ls -l ...

  3. Python数据结构:列表、字典、元组、集合

    列表:shoplist = ['apple', 'mango', 'carrot', 'banana']字典:di = {'a':123,'b':'something'}集合:jihe = {'app ...

  4. 分布式存储之MogileFS分布式文件系统简单应用

    一.分布式存储原理: 分布式存储系统,是将数据分散存储在多台独立的设备上.传统的网络存储系统采用集中的存储服务器存放所有数据,存储服务器成为系统性能的瓶颈,也是可靠性和安全性的焦点,不能满足大规模存储 ...

  5. vs2010中如何编写C语言程序

    File->New->Project 在打开的New Project对话框中最左侧一栏中选择Visual C++下面的CLR,之后在其右侧的区域中选择CLR Empty Applicati ...

  6. 文件下载—SSH框架文件下载

    1.准备下载的api组件 <dependency> <groupId>commons-io</groupId> <artifactId>commons- ...

  7. git使用多个SSH公钥信息

    常常在开发环境存在多个git库,比如官方的github.公司搭建的gitlab.自己的私人库等等多个git库,为了方便使用,git需要配置多个SSH公钥信息. 在centos7.5下,进入用户目录,以 ...

  8. SQL学习笔记五之MySQL索引原理与慢查询优化

    阅读目录 一 介绍 二 索引的原理 三 索引的数据结构 四 聚集索引与辅助索引 五 MySQL索引管理 六 测试索引 七 正确使用索引 八 联合索引与覆盖索引 九 查询优化神器-explain 十 慢 ...

  9. # 20145106 《Java程序设计》第3周学习总结

    教材学习内容总结 在本周的学习中,我看到了这样一句话:"使用java撰写程序几乎都是在使用对象(object),要产生对象必须先定义类(class),类是对象的设计图,对象是类的实例(ins ...

  10. 20145122 《Java程序设计》第4周学习总结

    教材学习内容总结 第六章 1.在java中,子类只能继承一个父类. 2.在java中,继承时使用extends关键字,private成员也会被继承. 3.检查多态语法逻辑是否正确,方式是从=号右边往左 ...