HDH 1264 Counting Squares （线段树+扫描线|暴力）

Counting Squares

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2113    Accepted Submission(s): 1056

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
  

 

Source

浙江工业大学第四届大学生程序设计竞赛 

 

题意：

给你几个矩形，算面积并。

POINT：

注意输入和输出方式，没有规定先输入左下和右上，都有可能，被这个坑的WA了好久。别的就和 HDU 1542一样。

还不需要离散化。

法2:

因为数据很小而且是整数，可以暴力。就是遍历一个坐标系，才100*100；

法1:

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <map>
#include <string.h>
#include <algorithm>
#include <vector>
#include <fstream>
using namespace std;
#define lt 2*x
#define rt 2*x+1
#define LL long long
const int N = 101*100;
int num=0;
int sum[N];
int cnt[N];
struct node
{
    int l,r;
    int h;
    int f;
}len[N];
bool cmd(node a,node b)
{
    return a.h<b.h;
}
void init()
{
    num=0;
    memset(sum,0,sizeof sum);
    memset(cnt,0,sizeof cnt);
    memset(len,0,sizeof len);
}
void pushup(int x,int l,int r)
{
    /* if(l==r&&cnt[x]) sum[x]=1;
     else if(cnt[x]) sum[x]=r-l;
     else
     sum[x]=sum[rt]+sum[lt];*/
    if(cnt[x]) sum[x]=r+1-l;
    else sum[x]=sum[lt]+sum[rt];
    
}
void add(int x,int l,int r,int ll,int rr,int f)
{
    if(ll<=l&&rr>=r)
    {
        cnt[x]+=f;
    }
    else
    {
        int mid=(l+r)>>1;
        if(ll<=mid) add(lt,l,mid,ll,rr,f);
        if(mid<rr) add(rt,mid+1,r,ll,rr,f);
    }
    pushup(x,l,r);
}
void gao()
{
    sort(len+1,len+1+num,cmd);
    int ans=0;
    for(int i=1;i<=num;i++)
    {
        add(1,0,1000,len[i].l,len[i].r-1,len[i].f);
        ans+=sum[1]*(len[i+1].h-len[i].h);
        
    }
    printf("%d\n",ans);
}
int main()
{
    int x1,x2,y1,y2;
    while(~scanf("%d %d %d %d",&x1,&y1,&x2,&y2))
    {
        if(x1==-1&&x2==-1&&y1==-1&&y2==-1)
        {
            gao();
            init();
        }
        else if(x1==-2&&x2==-2&&y1==-2&&y2==-2)
        {
            gao();
            break;
        }
        else
        {
            if(x1>x2) swap(x1, x2);
            if(y1>y2) swap(y1, y2);
            num++;
            len[num].f=1,len[num].l=x1,len[num].r=x2,len[num].h=y1;
            num++;
            len[num].f=-1,len[num].l=x1,len[num].r=x2,len[num].h=y2;
        }
    }
    
    return 0;
}

法2:

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
using namespace std;
int ans=0;
int flag[120][120];
int main()
{
    int x1,x2,y1,y2;
    while(~scanf("%d %d %d %d",&x1,&y1,&x2,&y2))
    {
        if(x1==-1&&x2==-1&&y1==-1&&y2==-1)
        {
            printf("%d\n",ans);
            memset(flag,0,sizeof flag);
            ans=0;
        }
        else if(x1==-2&&x2==-2&&y1==-2&&y2==-2)
        {
            printf("%d\n",ans);
            memset(flag,0,sizeof flag);
            ans=0;
            break;
        }
        else
        {
            for(int i=min(x1,x2);i<max(x2,x1);i++)
            {
                for(int j=min(y1,y2);j<max(y1,y2);j++)
                {
                    if(!flag[i][j])
                    {
                        ans++;
                        flag[i][j]=1;
                    }
                }
            }
        }
    }
    return 0;
}