POJ1151-扫面线+线段树+离散化//入门题

比较水的入门题

记录矩形竖边的x坐标，离散化排序。以被标记的边建树。

扫描线段树，查询线段树内被标记的边。遇到矩形的右边就删除此边

每一段的面积是查询结果乘边的横坐标之差，求和就是答案

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 1000;
int N,num,kase;
double savey[maxn*2];

struct line{
    double x,y1,y2;
    int flag;
    bool operator < (const struct line &t) const {return x < t.x;}
}Line[maxn];

struct Node{
    int l,r;
    double dl,dr;
    double len;
    int flag;
}SegTree[maxn*4];

void Build(int i,int l,int r)
{
    SegTree[i].l = l;
    SegTree[i].r = r;
    SegTree[i].flag = SegTree[i].len = 0;
    SegTree[i].dl = savey[l];
    SegTree[i].dr = savey[r];
    if(l +1 == r) return;
    int mid = (l+r)>>1;
    Build(i<<1,l,mid);
    Build(i<<1|1,mid,r);
}

void getlen(int t)
{
    if(SegTree[t].flag > 0)
    {
        SegTree[t].len = SegTree[t].dr - SegTree[t].dl;
        return ;
    }
    if(SegTree[t].l+1 == SegTree[t].r) SegTree[t].len = 0;
    else SegTree[t].len = SegTree[t<<1].len + SegTree[t<<1|1].len;
}

void Update(int i,line e)
{
    if(e.y1 == SegTree[i].dl && e.y2 == SegTree[i].dr)
    {
        SegTree[i].flag += e.flag;
        getlen(i);
        return;
    }
    if(e.y2 <= SegTree[i<<1].dr) Update(i<<1,e);
    else if(e.y1 >= SegTree[i<<1|1].dl)    Update(i<<1|1,e);
    else 
    {
        line temp = e;
        temp.y2 = SegTree[i<<1].dr;
        Update(i<<1,temp);
        temp = e;
        temp.y1 = SegTree[i<<1|1].dl;
        Update(i<<1|1,temp);
    }
    getlen(i);
}

int main()
{
    while(~scanf("%d",&N) && N)
    {
        kase++;
        num=1;
        double x1,x2,y1,y2;
        for(int i=1;i<=N;i++)
        {
            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
            Line[num].x = x1;
            Line[num].y1 = y1;
            Line[num].y2 = y2;
            Line[num].flag = 1;
            savey[num++]=y1;
            Line[num].x = x2;
            Line[num].y1 = y1;
            Line[num].y2 = y2;
            Line[num].flag = -1;
            savey[num++] = y2;
        }
        sort(Line+1,Line+num);
        sort(savey+1,savey+num);
        Build(1,1,num-1);
        Update(1,Line[1]);
        double ans = 0;
        for(int i=2;i<num;i++)
        {
            //printf("%f %f
",SegTree[1].len,Line[i].x-Line[i-1].x);
            ans += SegTree[1].len * (Line[i].x - Line[i-1].x);
            Update(1,Line[i]);
        }
        printf("Test case #%d
",kase);
        printf("Total explored area: %.2f

",ans);
    }    
    return 0;
}