UVA-201 Squares

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最新推荐文章于 2023-11-11 17:29:39 发布

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分类专栏：机智小水题算法竞赛入门经典练习题文章标签： UVA-201 Squares

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本文介绍了一种算法，用于解决一个特定的游戏板问题，即自动计算由一系列水平和垂直线条构成的不同大小的正方形数量。该算法适用于儿童游戏中的计数需求，并通过示例输入输出展示了其工作原理。

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Vjudge链接：

http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=19423

Squares

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

Submit Status

Description

Squares

A children's board game consists of a square array of dots that contains lines connecting some of the pairs of adjacent dots. One part of the game requires that the players count the number of squares of certain sizes that are formed by these lines. For example, in the figure shown below, there are 3 squares-2 of size 1 and 1 of size 2. (The ``size" of a square is the number of lines segments required to form a side.)

Your problem is to write a program that automates the process of counting all the possible squares.

Input

The input file represents a series of game boards. Each board consists of a description of a square array of n² dots (where 2 <= n <= 9) and some interconnecting horizontal and vertical lines. A record for a single board with n² dots and m interconnecting lines is formatted as follows:

Line 1: n the number of dots in a single row or column of the array

Line 2: m the number of interconnecting lines

Each of the next m lines are of one of two types:

H i j indicates a horizontal line in row i which connects

the dot in column j to the one to its right in column j + 1

V i j indicates a vertical line in column i which connects

the dot in row j to the one below in row j + 1

Information for each line begins in column 1. The end of input is indicated by end-of-file. The first record of the sample input below represents the board of the square above.

Output

For each record, label the corresponding output with ``Problem #1", ``Problem #2", and so forth. Output for a record consists of the number of squares of each size on the board, from the smallest to the largest. lf no squares of any size exist, your program should print an appropriate message indicating so. Separate output for successive input records by a line of asterisks between two blank lines, like in the sample below.

Sample Input

Sample Output

Problem #1

2 square (s) of size 1
1 square (s) of size 2

**********************************

Problem #2

No completed squares can be found.

Source

Root :: Competitive Programming 2: This increases the lower bound of Programming Contests. Again (Steven & Felix Halim) :: (Computational) Geometry :: Basic Geometry :: Quadrilaterals
Root :: Competitive Programming: Increasing the Lower Bound of Programming Contests (Steven & Felix Halim) :: Chapter 7. (Computational) Geometry :: Geometry Basics :: Rectangles

Root :: ACM-ICPC World Finals :: 1990 - Washington
Root :: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim) :: More Advanced Topics :: Problem Decomposition :: Two Components - Complete Search and Geometry
Root :: AOAPC II: Beginning Algorithm Contests (Second Edition) (Rujia Liu) :: Chapter 4. Functions and Recursion :: Exercises

题意：n个点组成的正方体，其中有m条边，H i j表示边（i，j）~（i，j+1），V j i表示边（i，j）~（i+1，j），问不同边长的子正方体各有多少个。

注意：

①V的ij是反的。

②输出的格式。

代码：

#include "stdio.h"
#include "string.h"
#include "iostream"
#include "stdlib.h"
#include "algorithm"
using namespace std;

int test=0,n,m,h[10][10],v[10][10],a,b,sq[10];
char r;

int jude(int x,int zsi,int zsj)
{
	for(int i=zsi;i<zsi+x;i++)
	{
		if(v[i][zsj]==0||v[i][zsj+x]==0)
		{
			return 0;
		}
	}
	for(int i=zsj;i<zsj+x;i++)
	{
		if(h[zsi][i]==0||h[zsi+x][i]==0)
		{
			return 0;
		}
	}
	return 1;
}

int main()
{
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		memset(h,0,sizeof(h));
		memset(v,0,sizeof(v));
		memset(sq,0,sizeof(sq));
		for(int i=1;i<=m;i++)
		{
			cin>>r>>a>>b;
			if(r=='H')
			{
				h[a][b]=1;
			}
			else
			{
				v[b][a]=1;
			}
		}
		for(int i=1;i<n;i++)
		{
			for(int j=1;j<=n-i;j++)
			{
				for(int w=1;w<=n-i;w++)
				{
					if(jude(i,j,w))
					{
						sq[i]++;
					}
				}
			}
		}
		if(test!=0)
		{
			printf("\n**********************************\n\n");
		}
		printf("Problem #%d\n\n",++test);
		int flag=0;
		for(int i=1;i<n;i++)
		{
			if(sq[i]!=0)
			{
				flag=1;
				printf("%d square (s) of size %d\n",sq[i],i);
			}
		}
		if(flag==0)
			printf("No completed squares can be found.\n");
	}
	return 0;
}