HDU 1542 Atlantis（线段树扫描线，面积并）

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题意：

给你n (n<=100)个矩阵，问你矩阵并后的面积。

解析：

http://www.cnblogs.com/kane0526/archive/2013/02/26/2934214.html
参考了这篇题解报告，终于学会了基本的线段树扫描线。

my code

#include <cstdio>#include <cstring>#include <algorithm>#define ls o<<1#define rs o<<1|1#define lson ls, L, M#define rson rs, M+1, Rusing namespace std;const int INF = 0x3f3f3f3f;const int maxn = 205;double X[maxn];int n, m, tot;struct Line {    double l, r, h;    int d;    Line() {}    Line(double x1, double x2, double h, int d) : l(x1), r(x2), h(h), d(d) {}    bool operator < (const Line& rhs) const {        return h < rhs.h;    }} line[maxn];double sumv[maxn<<2];int cover[maxn<<2];inline void pushUp(int o, int L, int R) {    if(cover[o]) sumv[o] = X[R+1] - X[L];    else if(L == R) sumv[o] = 0;    else sumv[o] = sumv[ls] + sumv[rs];}void build(int o, int L, int R) {    sumv[o] = cover[o] = 0;    if(L == R) return ;    int M = (L + R)/2;    build(lson);    build(rson);}void modify(int o, int L, int R, int ql, int qr, int d) {    if(ql <= L && R <= qr) {        cover[o] += d;        pushUp(o, L, R);        return ;    }    int M = (L + R)/2;    if(ql <= M) modify(lson, ql, qr, d);    if(qr > M) modify(rson, ql, qr, d);    pushUp(o, L, R);}int main() {    double x1, y1, x2, y2;    int cas = 1;    while(~scanf("%d", &n) && n) {        m = tot = 0;        for(int i = 0; i < n; i++) {            scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);            line[m] = Line(x1, x2, y1, 1);            X[m++] = x1;            line[m] = Line(x1, x2, y2, -1);            X[m++] = x2;        }        //discrete        sort(X, X+m);        tot = unique(X, X+m) - X;        sort(line, line+m);        build(1, 0, tot-1);        int ql, qr;        double ans = 0;        for(int i = 0; i < m; i++) {            ql = lower_bound(X, X+tot, line[i].l) - X;            qr = lower_bound(X, X+tot, line[i].r) - X - 1;            modify(1, 0, tot-1, ql, qr, line[i].d);            ans += sumv[1] * (line[i+1].h - line[i].h);        }        printf("Test case #%d\n", cas++);        printf("Total explored area: %.2lf\n\n", ans);    }    return 0;}

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