hdoj 1255 覆盖的面积【线段树 + 扫描线求重叠两次及以上的面积】

覆盖的面积

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4666    Accepted Submission(s): 2313

Problem Description

给定平面上若干矩形,求出被这些矩形覆盖过至少两次的区域的面积.

 

Input

输入数据的第一行是一个正整数T(1<=T<=100),代表测试数据的数量.每个测试数据的第一行是一个正整数N(1<=N<=1000),代表矩形的数量,然后是N行数据,每一行包含四个浮点数,代表平面上的一个矩形的左上角坐标和右下角坐标,矩形的上下边和X轴平行,左右边和Y轴平行.坐标的范围从0到100000.

注意:本题的输入数据较多,推荐使用scanf读入数据.

 

Output

对于每组测试数据,请计算出被这些矩形覆盖过至少两次的区域的面积.结果保留两位小数.

 

Sample Input

251 1 4 21 3 3 72 1.5 5 4.53.5 1.25 7.5 46 3 10 730 0 1 11 0 2 12 0 3 1

 

Sample Output

7.630.00

 

题意：给你n个矩形，求出重叠两次及以上的面积。

思路：理解矩形面积并就很好写了。设置sum1、sum2表示覆盖一、两次的面积，只需在Up时维护sum1和sum2就好了。

考虑节点区间被覆盖的程度cover >= 0，我们分开讨论

cover == 0，sum1 = [lson].sum1 + [rson].sum1, sum2 = [lson].sum2 + [rson].sum2;

cover == 1，sum2 = [lson].sum2 + [rson].sum2 + [lson].sum1 + [rson].sum1，sum1 = 线段len - sum2; 

cover >= 2，sum1 = 0, sum2 = 线段len。

AC代码：

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <set>#include <vector>#include <string>#define INF 0x3f3f3f3f#define eps 1e-8#define MAXN (10000+10)#define MAXM (200000+10)#define Ri(a) scanf("%d", &a)#define Rl(a) scanf("%lld", &a)#define Rf(a) scanf("%lf", &a)#define Rs(a) scanf("%s", a)#define Pi(a) printf("%d\n", (a))#define Pf(a) printf("%.2lf\n", (a))#define Pl(a) printf("%lld\n", (a))#define Ps(a) printf("%s\n", (a))#define W(a) while(a--)#define CLR(a, b) memset(a, (b), sizeof(a))#define MOD 1000000007#define LL long long#define lson o<<1, l, mid#define rson o<<1|1, mid+1, r#define ll o<<1#define rr o<<1|1#define PI acos(-1.0)using namespace std;struct Tree{    int l, r, len;    double sum1, sum2;    int cover;};Tree tree[MAXN<<2];struct Node{    double x1, x2, y;    int cover;};bool cmp(Node a, Node b){    return a.y < b.y;}Node num[MAXN];void Build(int o, int l, int r){    tree[o].l = l; tree[o].r = r;    tree[o].len = r-l+1;    tree[o].sum1 = tree[o].sum2 = tree[o].cover = 0;    if(l == r)        return ;    int mid = (l + r) >> 1;    Build(lson); Build(rson);}double rec[MAXN];int Find(int l, int r, double val){    while(r >= l)    {        int mid = (l + r) >> 1;        if(rec[mid] == val)            return mid;        else if(rec[mid] > val)            r = mid-1;        else            l = mid+1;    }}void PushUp(int o){    if(tree[o].cover >= 2)        tree[o].sum2 = rec[tree[o].r+1] - rec[tree[o].l], tree[o].sum1 = 0;    else if(tree[o].cover == 1)    {        if(tree[o].l == tree[o].r)            tree[o].sum2 = 0;        else            tree[o].sum2 = tree[ll].sum2 + tree[rr].sum2 + tree[ll].sum1 + tree[rr].sum1;        tree[o].sum1 = rec[tree[o].r+1] - rec[tree[o].l] - tree[o].sum2;    }    else    {        if(tree[o].l == tree[o].r)            tree[o].sum2 = tree[o].sum1 = 0;        else        {            tree[o].sum2 = tree[ll].sum2 + tree[rr].sum2;            tree[o].sum1 = tree[ll].sum1 + tree[rr].sum1;        }    }}void Update(int o, int L, int R, int v){    if(tree[o].l >= L && tree[o].r <= R)    {        tree[o].cover += v;        PushUp(o);        return ;    }    int mid = (tree[o].l + tree[o].r) >> 1;    if(R <= mid)        Update(ll, L, R, v);    else if(L > mid)        Update(rr, L, R, v);    else    {        Update(ll, L, mid, v);        Update(rr, mid+1, R, v);    }    PushUp(o);}int main(){    int t; Ri(t);    W(t)    {        int n; Ri(n);        int k = 0, len = 1;        for(int i = 0; i < n; i++)        {            double x1, y1, x2, y2;            Rf(x1); Rf(y1); Rf(x2); Rf(y2);            num[k].x1 = x1; num[k].x2 = x2;            num[k].y = y1;            num[k++].cover = 1;            rec[len++] = x1;            num[k].x1 = x1; num[k].x2 = x2;            num[k].y = y2;            num[k++].cover = -1;            rec[len++] = x2;        }        sort(num, num+k, cmp);        sort(rec+1, rec+len);        int R = 2;        for(int i = 2; i < len; i++)            if(rec[i] != rec[i-1])                rec[R++] = rec[i];        sort(rec, rec+R); R--;        Build(1, 1, R); double ans = 0;        for(int i = 0; i < k-1; i++)        {            int x = Find(1, R, num[i].x1);            int y = Find(1, R, num[i].x2);            if(i) ans += tree[1].sum2 * (num[i].y - num[i-1].y);            if(x <= y-1)                Update(1, x, y-1, num[i].cover);        }        printf("%.2lf\n", ans);    }    return 0;}