【线段树专题】poj1151

矩形面积并

线段树经典题目

这种题提供了一个新的思路，线段树的标记不一定需要下传。

就如同这个题我用biao这个域存了线段树的某个节点当前被多少线段覆盖，但是这个biao是不下传的。

然后这个题运用了扫描线的思想，把面积按照纵坐标分割。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<vector>
using namespace std;

int s_tot,tmp_tail,tail,tot,n;
double ans=0;

struct rec
{
    int l,r,biao,lch,rch;
    double sum,da;
}tree[20100];

struct rec2
{
    double y,x1,x2;
    int biao;
}segment[20100];

double tmp_q[20100],q[20100];

int cmp(rec2 x,rec2 y) {return x.y<y.y;}

void clear()
{
    memset(tree,0,sizeof(tree));
    memset(segment,0,sizeof(tree));
    memset(q,0,sizeof(q));
    memset(tmp_q,0,sizeof(q));
    s_tot=0;
    tmp_tail=0;
    tail=0;
    tot=0;
    ans=0;
}
void maketree(int l,int r)
{
    int now=++tot;
    tree[now].l=l,tree[now].r=r;
    if (l==r) 
    {
        tree[now].da=q[tree[now].l+1]-q[tree[now].l];
        return ;
    }
    int mid=(l+r)>>1;
    tree[now].lch=tot+1;maketree(l,mid);
    tree[now].rch=tot+1;maketree(mid+1,r);
    tree[now].da=tree[tree[now].lch].da+tree[tree[now].rch].da;
}

void Input()
{
    double x1,x2,y1,y2; 
    for (int i=1;i<=n;i++)
    {
        scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
        s_tot++;
        segment[s_tot].y=y1;
        segment[s_tot].x1=x1;
        segment[s_tot].x2=x2;
        segment[s_tot].biao=1;
        s_tot++;
        segment[s_tot].y=y2;
        segment[s_tot].x1=x1;
        segment[s_tot].x2=x2;
        segment[s_tot].biao=-1;
        tmp_q[++tmp_tail]=x1;
        tmp_q[++tmp_tail]=x2;
    }
    sort(tmp_q+1,tmp_q+tmp_tail+1);
    sort(segment+1,segment+s_tot+1,cmp);
    tmp_q[0]=300000;
    for (int i=1;i<=s_tot;i++)
        if (tmp_q[i]!=tmp_q[i-1])
            q[++tail]=tmp_q[i];
    maketree(1,tail-1);
}
void change(int x,int l,int r,int w)
{
    if (tree[x].l>=l&&tree[x].r<=r)
    {
        tree[x].biao+=w;
        if (tree[x].biao!=0) tree[x].sum=tree[x].da;
        else
        {
            if (tree[x].l==tree[x].r) tree[x].sum=0;
            else tree[x].sum=tree[tree[x].lch].sum+tree[tree[x].rch].sum;
        }
        return;
    }
    int mid=(tree[x].l+tree[x].r)>>1;
    if (l<=mid) change(tree[x].lch,l,r,w);
    if (r>mid) change(tree[x].rch,l,r,w);
    if (tree[x].biao!=0) tree[x].sum=tree[x].da;
    else
    {
        if (tree[x].l==tree[x].r) tree[x].sum=0;
        else tree[x].sum=tree[tree[x].lch].sum+tree[tree[x].rch].sum;
    }
}
void work()
{
    int l,r;
    segment[0].y=200100;
    for (int i=1;i<=s_tot;i++)
    {
        if (segment[i].y!=segment[i-1].y)
            ans+=(segment[i].y-segment[i-1].y)*tree[1].sum;
        l=lower_bound(q+1,q+tail+1,segment[i].x1)-q;
        r=lower_bound(q+1,q+tail+1,segment[i].x2)-q;    
        change(1,l,r-1,segment[i].biao);
    }
    printf("Total explored area: %.2f\n\n",ans);
}
int main()
{
    freopen("poj1151.in","r",stdin);
    freopen("poj1151.out","w",stdout);
    int ci=0;
    while (scanf("%d",&n)&&n!=0)
    {
        ci++;
        printf("Test case #%d\n",ci);
        clear();
        Input();
        work();
    }
    return 0;
}

奇怪的是交c++会RE但是交G++就能过，目测是中间用的lower_bound的问题？