HDU 1255 覆盖的面积 【线段树】【矩形面积交（扫描线，离散化）】

覆盖的面积

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7580    Accepted Submission(s): 3828

 

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

2 5 1 1 4 2 1 3 3 7 2 1.5 5 4.5 3.5 1.25 7.5 4 6 3 10 7 3 0 0 1 1 1 0 2 1 2 0 3 1

 

 

Sample Output

7.63 0.00

 

#include
using namespace std;
#define db double
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
const int MAX = 100005;
struct node{
    db x, y, h;
    int flag;
    node() {}
    node(db x, db y, db h, int flag) : x(x), y(y), h(h), flag(flag) {}
    bool operator < (const node &a) const{
        return h < a.h;
    }
} e[MAX];
db X[MAX];
db lz[MAX];
db len[MAX], Len[MAX];
void pushup(int rt, int l, int r){
    if(lz[rt])
        len[rt] = X[r + 1] - X[l];
    else if(l == r) len[rt] = 0;
    else len[rt] = len[rt << 1] + len[rt << 1 | 1];
    if(lz[rt] > 1) Len[rt] = X[r + 1] - X[l];
    else if(l == r) Len[rt] = 0;
    else if(lz[rt] == 1)
        Len[rt] = len[rt << 1] + len[rt << 1 | 1];
    else Len[rt] = Len[rt << 1] + Len[rt << 1 | 1];
}
void update(int rt, int l, int r, int x, int y, int v){
    if(x <= l && r <= y){
        lz[rt] += v;
        pushup(rt, l, r);
        return;
    }
    int mid = (l + r) >> 1;
    if(x <= mid) update(lson, x, y, v);
    if(mid < y) update(rson, x, y, v);
    pushup(rt, l, r);
}
int main(){
    int N;
    scanf("%d", &N);
    while(N--){
        int n;
        memset(X, 0 ,sizeof X);
        memset(lz, 0, sizeof lz);
        memset(len, 0, sizeof len);
        memset(Len, 0, sizeof Len);
        int cnt = 0;
        scanf("%d", &n);
        for(int i = 0; i < n; i++){
            db a, b, c, d;
            scanf("%lf%lf%lf%lf", &a, &b, &c, &d);
            e[cnt] = node(a, c, b, 1);
            X[cnt++] = a;
            e[cnt] = node(a, c, d, -1);
            X[cnt++] = c;
        }
        sort(X, X + cnt), sort(e, e + cnt);
        db ans = 0.0;
        int num = unique(X, X + cnt) - X;
        for(int i = 0; i < cnt - 1; i++){
            int l = lower_bound(X, X + num, e[i].x) - X;
            int r = lower_bound(X, X + num, e[i].y) - X - 1;
            update(1, 0, num - 1, l, r, e[i].flag);
            ans += Len[1] * (e[i + 1].h - e[i].h);
        }
        printf("%.2f\n", ans);
    }
    return 0;
}