zoukankan      html  css  js  c++  java
  • 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;
    }
  • 相关阅读:
    自己写的DBHelper感慨颇深
    23种设计模式:观察者模式,第一次对观察者模式理解的这么通透。
    自己用心写的 存储过程分页 给自己的平台用
    开篇成长的开始[废话一把]
    C# 中i++在ref参数调用下的有趣现象
    点点滴滴的成长[2011111]:理解C#修饰符
    点点滴滴的成长[2011114]:自定义config文件
    扩展方法在Json字符串转化中的应用
    jquery学习二:jquery性能优化
    javascript系列1:函数
  • 原文地址:https://www.cnblogs.com/gpsx/p/5185573.html
Copyright © 2011-2022 走看看