hdu 1255 覆盖的面积(求覆盖至少两次以上的面积)

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覆盖的面积

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4823    Accepted Submission(s): 2398

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

/*hdu 1255 覆盖的面积(求覆盖至少两次以上的面积)主要是在push_up那出现了问题设定len是覆盖两次以上的长度，len1是覆盖1次以上的长度那么有以下几种情况：1.sum>1 则说当前区间被覆盖至少两次2.l == r  直接为03.当前区间一次都没被覆盖过 lson.len + rson.len4.当前区间被覆盖过一次sum=1,那么  lson.len1 + rson.len1，即左右孩子中被  //最开始没有考虑到这个情况TAT至少覆盖过一次的和hhh-2016-03-27 18:07:36*/#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <functional>using namespace std;#define lson  (i<<1)#define rson  ((i<<1)|1)typedef long long ll;const int maxn = 50050;struct node{    int l,r,sum;    double len;    double len1;    int mid()    {        return (l+r)>>1;    }} tree[maxn*5];double hs[maxn];int tot,m;void push_up(int i){    if(tree[i].sum)        tree[i].len1 = hs[tree[i].r+1]-hs[tree[i].l];    else if(tree[i].l == tree[i].r)        tree[i].len1 = 0;    else        tree[i].len1 = tree[lson].len1+tree[rson].len1;    if(tree[i].sum > 1)        tree[i].len = hs[tree[i].r+1]-hs[tree[i].l];    else if(tree[i].l == tree[i].r)        tree[i].len = 0;    else if(tree[i].sum == 1)        tree[i].len = tree[lson].len1 + tree[rson].len1;    else        tree[i].len = tree[lson].len+tree[rson].len;}void build(int i,int l,int r){    tree[i].l = l,tree[i].r = r;    tree[i].sum=tree[i].len= 0;    tree[i].len1 = 0;    if(l ==r )    {        return ;    }    int mid=tree[i].mid();    build(lson,l,mid);    build(rson,mid+1,r);    push_up(i);}void push_down(int i){}void Insert(int i,int l,int r,int val){    if(tree[i].l >= l && tree[i].r <= r)    {        tree[i].sum += val;        push_up(i);        return;    }    push_down(i);    int mid = tree[i].mid();    if(l <= mid)        Insert(lson,l,r,val);    if(r > mid)        Insert(rson,l,r,val);    push_up(i);}struct edge{    double l,r,high;    int va;    edge() {};    edge(double ll,double rr,double h,int v):l(ll),r(rr),high(h),va(v) {}} tx[maxn];bool cmp(edge a,edge b){    if(a.high != b.high)        return a.high < b.high;    else        return a.va > b.va;}int fin(double x){    int l = 0,r = m-1;    while(l <= r)    {        int mid = (l+r)>>1;        if(hs[mid] == x)            return mid;        else if(hs[mid] < x)            l = mid+1;        else            r = mid-1;    }}int main(){    int n,x,q,t;    int cas =1;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        tot = 0;        double x1,x2,y1,y2;        for(int i = 1; i <= n; i++)        {            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);            hs[tot] = x1;            tx[tot++] = edge(x1,x2,y1,1);            hs[tot] = x2;            tx[tot++] = edge(x1,x2,y2,-1);        }        sort(hs,hs+tot);        sort(tx,tx+tot,cmp);        m = 1;        for(int i = 1; i < tot; i++)            if(hs[i] != hs[i-1])                hs[m++] = hs[i];        build(1,0,m-1);        double ans = 0;        for(int i = 0;i < tot;i++)        {            int l = fin(tx[i].l);            int r = fin(tx[i].r)-1;            //cout << tx[i].l <<" " << tx[i].r <<" "<<endl;            Insert(1,l,r,tx[i].va);            ans += tree[1].len*(tx[i+1].high-tx[i].high);        }        printf("%.2f\n",ans);    }    return 0;}