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POJ 1151 / HDU 1542 Atlantis 线段树求矩形面积并

题意:给出矩形两对角点坐标,求矩形面积并。

解法:线段树+离散化。

每加入一个矩形,将两个y值加入yy数组以待离散化,将左边界cover值置为1,右边界置为2,离散后建立的线段树其实是以y值建的树,线段树维护两个值:cover和len,cover表示该线段区间目前被覆盖的线段数目,len表示当前已覆盖的线段长度(化为离散前的真值),每次加入一条线段,将其y_low,y_high之间的区间染上line[i].cover,再以tree[1].len乘以接下来的线段的x坐标减去当前x坐标,即计算了一部分面积。

如图情况,将会计算三次面积:

代码:

#include <iostream>#include <cmath>#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <algorithm>#define eps 1e-8using namespace std;#define N 307struct node{    int cov;    double len;}tree[8*N];struct Line{    double y1,y2,x;    int cov;}line[N];double yy[N];int cmp(Line ka,Line kb){    return ka.x < kb.x;}int bsearch(int l,int r,double x){    while(l <= r)    {        int mid = (l+r)/2;        if(fabs(yy[mid]-x) < eps)            return mid;        if(x > yy[mid])            l = mid+1;        else            r = mid-1;    }    return l;}void build(int l,int r,int rt){    tree[rt].cov = 0;    tree[rt].len = 0;    if(l == r-1) return;    int mid = (l+r)/2;    build(l,mid,2*rt);    build(mid,r,2*rt+1);}void pushup(int l,int r,int rt){    if(tree[rt].cov)        tree[rt].len = yy[r]-yy[l];    else if(l+1 == r)        tree[rt].len = 0;    else        tree[rt].len = tree[2*rt].len + tree[2*rt+1].len;}void update(int l,int r,int aa,int bb,int cover,int rt){    if(aa <= l && bb >= r)    {        tree[rt].cov += cover;        pushup(l,r,rt);        return;    }    if(l+1 == r) return;    int mid = (l+r)/2;    if(aa <= mid)        update(l,mid,aa,bb,cover,2*rt);    if(bb > mid)        update(mid,r,aa,bb,cover,2*rt+1);    pushup(l,r,rt);}int main(){    int n,m,cs = 1,i,j;    double x1,y1,x2,y2;    while(scanf("%d",&n)!=EOF && n)    {        m = 1;        for(i=0;i<n;i++)        {            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);            line[m].x = x1,line[m].y1 = y1,line[m].y2 = y2,line[m].cov = 1,yy[m++] = y1;            line[m].x = x2,line[m].y1 = y1,line[m].y2 = y2,line[m].cov = -1,yy[m++] = y2;        }        m--;        sort(yy+1,yy+m+1);        int cnt = 2;        for(i=2;i<=m;i++)        {            if(yy[i] != yy[i-1])                yy[cnt++] = yy[i];        }        cnt--;        build(1,cnt,1);        sort(line+1,line+m+1,cmp);        double ans = 0.0;        printf("Test case #%d\n",cs++);        for(i=1;i<m;i++)        {            int L = bsearch(1,cnt,line[i].y1);            int R = bsearch(1,cnt,line[i].y2);            update(1,cnt,L,R,line[i].cov,1);            ans += tree[1].len*(line[i+1].x-line[i].x);        }        printf("Total explored area: %0.2lf\n\n",ans);    }    return 0;}
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POJ 1151 / HDU 1542 Atlantis 线段树求矩形面积并