(简单) 搜索 HOJ 1085 Finding Rectangles

Finding Rectangles

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Source : ACM ICPC Mid-Central USA 1998Time limit : 1 secMemory limit : 32 M

Submitted : 258, Accepted : 118

Consider the point sets in figures 1a, 2a, and 3a. Using only those points as vertices, figures 1b, 2b, and 3b show all the rectangles that can be formed with horizontal and vertical sides. No rectangles can be formed from the points in figure 4.

"""

Your task is to write a program that can find all rectangles that can be formed from a given set of points. The example input and output given below correspond to the figures above.

Input

The input contains one or more point sets, followed by a line containing the number 0 that signals the end of the file. Each point set begins with a line containing n, the number of points, and is followed by n lines that describe the points. Each point description contains a capital letter that is the label of the point, then a space, the horizontal coordinate, a space, and the vertical coordinate. Within each set, points labels occur in alphabetical order.

Note that since each point is labelled with a capital letter there can be at most 26 points. All coordinates are nonnegative integers less than 50. Points within a set are unique.

Output

The output for each point set starts with ``Point set ", followed by the number of the point set and a colon. If there are no rectangles, `` No rectangles" appears after the colon. If there are rectangles, they are listed starting on the next line. A blank precedes each rectangle. Each rectangle is given by its vertex labels, in clockwise order from the upper left, so the order is upper left, upper right, lower right, lower left. The rectangles are listed ten per line, except for the last line, where there may be as few as one. The rectangles are listed in alphabetical order.

Sample Input

7A 1 1 B 2 1 C 3 1 D 2 3 E 3 3 F 1 4 G 3 4 8 B 1 1 D 2 1 F 4 1 J 4 4 L 2 4M 2 3N 4 3 P 1 2 12A 1 5B 2 5C 1 4D 2 4E 1 3F 2 3G 1 2H 2 2I 1 1J 2 1K 1 0L 2 05B 1 1D 2 1L 2 4N 2 3P 1 20

Sample Output

Point set 1: DECB FGCAPoint set 2: LJFD LJNM MNFDPoint set 3: ABDC ABFE ABHG ABJI ABLK CDFE CDHG CDJI CDLK EFHG EFJI EFLK GHJI GHLK IJLKPoint set 4: No rectangles

题意：给一系列的点，找出所有的矩形

思路：数据量不大，直接枚举，检查是否是矩形就行了 ， 最后注意输出的格式

代码：

#include<iostream>

#include<cstdio>

#include<string.h>

#include<algorithm>

#include<string>

#include<deque>

#include<queue>

#include<math.h>

#include<vector>

#include<map>

#include<stack>

#include<set>

using namespace std;

#define MAX 100+10

#define MOD 99997

const int inf = 0xfffffff;

struct Point

{

int x,y;

}point[MAX];

bool operator<(const Point& p1,const Point& p2)

{

if (p1.x==p2.x) return p1.y<p2.y;

return p1.x<p2.x;

}

string ans[MAX];

char match[MAX][MAX];

int n;

int ans_sz;

bool flag;

void check_output(stack<Point> rect)

{

vector<Point> tem;

while (rect.size())

{

tem.push_back(rect.top());

rect.pop();

}

sort(tem.begin(),tem.end());

if (tem[0].x!=tem[1].x) return;

if (tem[2].x!=tem[3].x) return;

if (tem[2].x==tem[0].x) return;

if (tem[0].y>=tem[1].y) return;

if (tem[0].y!=tem[2].y) return;

if (tem[1].y!=tem[3].y) return;

ans[ans_sz] = "";

ans[ans_sz] += match[tem[1].x][tem[1].y];

ans[ans_sz] += match[tem[3].x][tem[3].y];

ans[ans_sz] += match[tem[2].x][tem[2].y];

ans[ans_sz++] += match[tem[0].x][tem[0].y];

flag = true;

}

void dfs(int cur,stack<Point> rect)

{

if (rect.size()==4)

{

check_output(rect);

return;

}

for (int i = cur ; i < n ; ++i)

{

rect.push(point[i]);

dfs(i+1,rect);

rect.pop();

}

}

int main()

{

int cas = 0;

while (scanf("%d",&n),n)

{

++cas;

for (int i = 0 ; i < n ; ++i)

{

char ch;

scanf(" %c%d%d",&ch,&point[i].x,&point[i].y);

match[point[i].x][point[i].y] = ch;

}

stack<Point> emty;

printf("Point set %d:",cas);

ans_sz = 0;

flag = false;

dfs(0,emty);

if (ans_sz==0)

{

printf(" No rectangles\n");

continue;

}

sort(ans,ans+ans_sz);

int cnt = 0;

printf("\n");

for (int i = 0 ; i < ans_sz ; ++i)

{

printf(" %s",ans[i].c_str());

++cnt;

if (cnt%10==0)

printf("\n");

}

if (cnt%10) printf("\n");

}

}