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POJ 1151 Atlantis

题目大意:

给出n个矩形,形式是左下点和右上点。求它们的面积并。



解题思路:

扫描线算法,对Y进行扫描,线段树查询Y轴扫描某段距离后X轴一共有多长的距离有边,并计算面积。



下面是代码:

#include <set>
#include <map>
#include <queue>
//#include <math.h>
#include <vector>
#include <string>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>

#define eps 1e-8
#define pi acos(-1.0)
#define inf 107374182
#define inf64 1152921504606846976
#define lc l,m,tr<<1
#define rc m + 1,r,tr<<1|1
#define iabs(x)  ((x) > 0 ? (x) : -(x))
#define clear1(A, X, SIZE) memset(A, X, sizeof(A[0]) * (SIZE))
#define clearall(A, X) memset(A, X, sizeof(A))
#define memcopy1(A , X, SIZE) memcpy(A , X ,sizeof(X[0])*(SIZE))
#define memcopyall(A, X) memcpy(A , X ,sizeof(X))
#define max( x, y )  ( ((x) > (y)) ? (x) : (y) )
#define min( x, y )  ( ((x) < (y)) ? (x) : (y) )


using namespace std;

struct node2
{
    int num;
    double y,l,r;
} edge[1000];

double tempx[1000],binx[1000];
double x1,x2,y1,y2,ans;
int cntx,n;

bool cmp(node2 a,node2 b)
{
    return a.y<b.y;
}

int binnum(double num)
{
    int ll=0,m,rr=cntx-1;
    while(rr>ll)
    {
        m=(ll+rr)>>1;
        if(binx[m]==num)return m;
        else if(binx[m]<num)ll=m+1;
        else rr=m-1;
    }
    return ll;
}

struct node1
{
    double disnow;
    int cnt;
} node[205<<2];

inline void PushUp(int l,int r,int tr)
{
    if(node[tr].cnt)node[tr].disnow=binx[r+1]-binx[l];
    else if(l==r)node[tr].disnow=0;
    else node[tr].disnow=node[tr<<1].disnow+node[tr<<1|1].disnow;
}

void update(int L,int R,int num,int l,int r,int tr)
{
    if(L<=l&&r<=R)
    {
        node[tr].cnt+=num;
        PushUp(l,r,tr);
        return ;
    }
    int m=(l+r)>>1;
    if(L<=m)update(L,R,num,l,m,tr<<1);
    if(m<R)update(L,R,num,m+1,r,tr<<1|1);
    PushUp(l,r,tr);
}
int main()
{
    int case1=1;
    while(scanf("%d",&n),n)
    {
        for(int i=0; i<n; i++)
        {
            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
            tempx[2*i]=x1;
            tempx[2*i+1]=x2;
            edge[2*i].l=x1;
            edge[2*i].r=x2;
            edge[2*i+1]=edge[2*i];
            edge[2*i].y=y1;
            edge[2*i+1].y=y2;
            edge[2*i].num=1;
            edge[2*i+1].num=-1;
        }
        sort(tempx,tempx+2*n);
        binx[0]=tempx[0];
        cntx=1;
        for(int i=1; i<2*n; i++)
        {
            if(tempx[i]!=binx[cntx-1])
            {
                binx[cntx++]=tempx[i];
            }
        }
        sort(edge,edge+2*n,cmp);
        clearall(node,0);
        ans=0;
        x1=binnum(edge[0].l);
        x2=binnum(edge[0].r);
        x2--;
        update(x1,x2,edge[0].num,0,cntx-2,1);
        y1=edge[0].y;
        for(int i=1; i<2*n; i++)
        {
            ans+=node[1].disnow*(edge[i].y-y1);
            y1=edge[i].y;
            x1=binnum(edge[i].l);
            x2=binnum(edge[i].r);
            x2--;
            update(x1,x2,edge[i].num,0,cntx-2,1);
        }
        printf("Test case #%d\n",case1++);
        printf("Total explored area: %.2lf\n\n",ans);
    }
    return 0;
}