HDH 1264 Counting Squares （線段樹+掃描線|暴力）

Mr_Treeeee發表於2020-04-06

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



浙江工業大學第四屆大學生程式設計競賽 



 


題意：


給你幾個矩形，算面積並。


POINT：


注意輸入和輸出方式，沒有規定先輸入左下和右上，都有可能，被這個坑的WA了好久。別的就和HDU 1542一樣。


還不需要離散化。


法2:


因為資料很小而且是整數，可以暴力。就是遍歷一個座標系，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;
}




                                    




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