HDU 1264 Counting Squares

Counting Squares

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 105    Accepted Submission(s): 59

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

 

 

Recommend

JGShining

题目的意思很简单，给定几个长方形，叫你求这些长方形的并起来的面积

本人使用的是离散化+线段树解决的

首先把所有的X坐标排序进行离散化

然后枚举每一个区间(X[i],X[i+1]);

再扫描所有的长方形，如果有长方形能覆盖(X[i],X[i+1]),那么在线段树中插入线段[Y1,Y2] (Y2>Y1)

之后统计该区间内的线段总长度L，ans+=L*(X[i+1]-X[i])

#include<iostream>
#include<algorithm>
using namespace std;
struct Seg
{
    int l,r,val;
}seg[305];
struct Line
{
    int x1,x2,y1,y2;
}line[105];
int X[205],len,llen,MV=0;
int ans,val=0;
void mkseg(int l,int r,int x)
{
    int mid=(l+r)>>1;
    seg[x].l=l;
    seg[x].r=r;
    seg[x].val=0;
    MV=MV>x?MV:x;
    if(r-l==1)
        return;
    mkseg(l,mid,2*x);
    mkseg(mid,r,2*x+1);
}
void ins(int l,int r,int x)
{
    if(x>MV)return;
    if(l<=seg[x].l&&seg[x].r<=r)
    {
        seg[x].val=val;
        return;
    }
    int mid=(seg[x].l+seg[x].r)>>1;
    if(r<=mid)
        ins(l,r,2*x);
    else if(l>=mid)
        ins(l,r,2*x+1);
    else ins(l,mid,2*x),ins(mid,r,2*x+1);
}
void gans(int x)
{
    if(x>MV)return;
    if(seg[x].val==val)
    {
        ans+=(seg[x].r-seg[x].l);
        return;
    }
    gans(2*x);
    gans(2*x+1);
}
int main()
{
    int a,b,c,d,i,j,area;
    bool end=false;
    mkseg(0,100,1);
    while(scanf("%d%d%d%d",&X[0],&a,&X[1],&b)!=EOF)
    {
        if(X[0]==-1)
        {
            puts("0");
            continue;
        }
        if(X[0]==-2)return 0;
        area=0;
        len=2;
        if(X[0]>X[1])swap(X[0],X[1]);
        line[0].x1=X[0];
        line[0].x2=X[1];
        if(a>b){a^=b;b^=a;a^=b;}
        line[0].y1=a;
        line[0].y2=b;
        llen=1;
        while(scanf("%d%d%d%d",&a,&b,&c,&d)!=EOF)
        {
            if(d==-1)break;
            if(d==-2){end=true;break;}
            if(b>d){b^=d;d^=b;b^=d;}
            if(a>c){a^=c;c^=a;a^=c;}
            X[len]=a;
            X[len+1]=c;
            line[llen].x1=a;
            line[llen].x2=c;
            line[llen].y1=b;
            line[llen].y2=d;
            llen++;
            len+=2;
        }
        sort(X,X+len);
        for(i=0;i<len-1;i++)
        {    //X坐标的区间段...
            if(X[i]==X[i+1])continue;
            ++val;
            for(j=0;j<llen;j++)
            {
                if(line[j].x1<=X[i]&&line[j].x2>=X[i+1])ins(line[j].y1,line[j].y2,1);
            }
            ans=0;
            gans(1);
            area+=ans*(X[i+1]-X[i]);
        }
        printf("%d/n",area);
        if(end)return 0;
    }
    return 0;
}

下面是今天写的离散化的，复杂度是o(n^3)居然0ms....

#include<iostream>
#include<algorithm>
using namespace std;
int X[2002],Y[2002],Len,Rlen;
struct Rec
{
    int xlow,xhigh,ylow,yhigh;
}R[1001];
int main()
{
    int ans;
    int i,j,k;
    int x1,y1,x2,y2;
    bool flag=false;
    while(!flag)
    {
        Rlen=1;
        Len=1;
        while(scanf("%d%d%d%d",&x1,&y1,&x2,&y2)!=EOF)
        {
            if(x1==-1&&x2==-1&&y1==-1&&y2==-1)break;
            if(x1==-2&&x2==-2&&y1==-2&&y2==-2){flag=true;break;}
            if(x1>x2)swap(x1,x2);
            if(y1>y2)swap(y1,y2);
            X[Len]=x1;
            X[Len+1]=x2;
            Y[Len]=y1;
            Y[Len+1]=y2;
            R[Rlen].xlow=x1;
            R[Rlen].xhigh=x2;
            R[Rlen].ylow=y1;
            R[Rlen].yhigh=y2;
            Rlen++;
            Len+=2;
        }
        ans=0;
        sort(X+1,X+Len);
        sort(Y+1,Y+Len);
        for(i=1;i<Len-1;i++)
        {
            if(X[i]==X[i+1])continue;
            for(j=1;j<Len-1;j++)
            {
                if(Y[j]==Y[j+1])continue;
                for(k=1;k<Rlen;k++)
                {
                    if(R[k].xlow<=X[i]&&R[k].xhigh>=X[i+1]&&R[k].ylow<=Y[j]&&R[k].yhigh>=Y[j+1])
                    {
                        ans+=(X[i+1]-X[i])*(Y[j+1]-Y[j]);
                        break;
                    }
                }
            }
        }
        printf("%d/n",ans);
    }
    return 0;
}