HDU 1264 Counting Squares(模拟)

题目链接

Problem Description

Your input is a series of rectangles, one per line. Each rectangle is specified as two points(X,Y) that specify the opposite corners of a rectangle. All coordinates will be integers in the range 0 to 100. For example, the line
5 8 7 10
specifies the rectangle who's corners are(5,8),(7,8),(7,10),(5,10).
If drawn on graph paper, that rectangle would cover four squares. Your job is to count the number of unit(i.e.,1*1) squares that are covered by any one of the rectangles given as input. Any square covered by more than one rectangle should only be counted once.

 

Input

The input format is a series of lines, each containing 4 integers. Four -1's are used to separate problems, and four -2's are used to end the last problem. Otherwise, the numbers are the x-ycoordinates of two points that are opposite corners of a rectangle.

 

Output

Your output should be the number of squares covered by each set of rectangles. Each number should be printed on a separate line.

 

Sample Input

5 8 7 10

6 9 7 8

6 8 8 11

-1 -1 -1 -1

0 0 100 100

50 75 12 90

39 42 57 73

-2 -2 -2 -2

 

Sample Output

8

10000

 题解：水题，因为数字范围不大，可以用二维数组保存每个1x1的块是否被访问过。

#include <cstdio>
#include <iostream>
#include <string>
#include <sstream>
#include <cstring>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <map>
#define PI acos(-1.0)
#define ms(a) memset(a,0,sizeof(a))
#define msp memset(mp,0,sizeof(mp))
#define msv memset(vis,0,sizeof(vis))
using namespace std;
//#define LOCAL
int mp[120][120];
int main()
{
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
#endif // LOCAL
    ios::sync_with_stdio(false);
    int x1,y1,x2,y2;
    int cnt=0;
    msp;
    while(cin>>x1>>y1>>x2>>y2)
    {
        if(x1==-1){printf("%d
",cnt);
        msp,cnt=0;}
        if(x1==-2){
        printf("%d
",cnt);break;}
        if(x1>x2)swap(x1,x2);
        if(y1>y2)swap(y1,y2);
        for(int i=x1;i<x2;i++)
        for(int j=y1;j<y2;j++)
        if(mp[i][j])continue;
        else {cnt++,mp[i][j]=1;}
    }
    return 0;
}