[Codevs] 矩形面积求并

http://codevs.cn/problem/3044/

线段树扫描线矩形面积求并

基本思路就是将每个矩形的长（平行于x轴的边）投影到线段树上

下边+1，上边-1；

然后根据线段树的权值和与相邻两条边的差值（高度差）求出相应的矩形面积

#include <bits/stdc++.h>

using namespace std;
const int N = 1100;

#define DB double

int F[N << 2];
DB W[N << 2], X[N];
struct Node {
    DB x_1, x_2, h; int how;
    Node() {};
    Node(DB x_1_, DB x_2_, DB h_, int how_) {x_1 = x_1_; x_2 = x_2_; h = h_; how = how_;}
}A[N];

bool cmp(Node a, Node b) {return a.h < b.h;}

inline int Find(DB num, int n) {
    int L = 1, R = n, Mid;
    while(L <= R) {
        Mid = (L + R) >> 1;
        if(X[Mid] == num) return Mid;
        else if(X[Mid] < num) L = Mid + 1;
        else R = Mid - 1;
    }
}

#define lson jd << 1
#define rson jd << 1 | 1 

void Pushup(int jd, int l, int r) {
    if(F[jd]) W[jd] = X[r + 1] - X[l];
    else if(l == r) W[jd] = 0;
    else W[jd] = W[lson] + W[rson];
} 

void Sec_G(int l, int r, int jd, int x, int y, int yj) {
    if(x <= l && r <= y) {
        F[jd] += yj;
        Pushup(jd, l, r);
        return ;
    }
    int mid = (l + r) >> 1;
    if(x <= mid) Sec_G(l, mid, lson, x, y, yj);
    if(y > mid)  Sec_G(mid + 1, r, rson, x, y, yj);
    Pushup(jd, l, r);
}

int main() {
    while(1) {
        int n;
        cin >> n;
        int js = 0;
        if(!n) break;
        memset(F, 0, sizeof F);
        memset(W, 0, sizeof W);
        for(int i = 1; i <= n; i ++) {
            DB x_1, x_2, y_1, y_2;                                                                                 
            scanf("%lf%lf%lf%lf", &x_1, &y_1, &x_2, &y_2);
            A[++ js] = Node(x_1, x_2, y_1, 1);
            X[js] = x_1;
            A[++ js] = Node(x_1, x_2, y_2, -1);
            X[js] = x_2;
        }
        sort(X + 1, X + js + 1);
        sort(A + 1, A + js + 1, cmp);
        int k = 1;
        for(int i = 2; i <= js; i ++) if(X[i] != X[i + 1]) X[++ k] = X[i];
        DB Answer = 0;
        for(int i = 1; i < js;  i ++) {
            int l = Find(A[i].x_1, k), r = Find(A[i].x_2, k) - 1;
            Sec_G(1, k, 1, l, r, A[i].how);
            Answer += W[1] * (A[i + 1].h - A[i].h);
        }
        printf("%.2lf
", Answer);
    }
    return 0;
}