POJ 1151 Atlantis

来源：互联网发布：西安丝路软件编辑：程序博客网时间：2024/05/22 15:13

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;}

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