HDU-2461 Rectangles 线段树，矩形面积并

首先申明此方法POJ超时，HDU压线过，优化版见http://www.cnblogs.com/Lyush/archive/2012/07/28/2613516.html

线段树的写法与上面链接中的离散化版本的想法是相近的，只不过这里仅仅是通过线段树来保留某一x区域的多个矩形的面积并。

代码如下：

#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <map>
#include <algorithm>
using namespace std;

int N, M, Q, cnt;

map<int,int>mp;

struct Rectangle
{
    int x1, y1, x2, y2;
}e[25];

struct High
{
    int y;
    bool operator < (High temp) const
    {
        return y < temp.y;
    }
    bool operator == (High temp) const
    {
        return y == temp.y;
    }
}H[50];

struct xLine
{
    int x, y1, y2, sign;
    bool operator < (xLine temp) const
    {
        return x < temp.x;
    }
}Line[50];

struct Node
{
    int l, r, lazy;
}s[200];

void build(int p, int l, int r)
{
    s[p].l = l, s[p].r = r;
    s[p].lazy = 0;  // 初始化为被覆盖了0次
    if (l != r) {
        int mid = (l + r) >> 1;
        build(p<<1, l, mid);
        build(p<<1|1, mid+1, r);
    }
}

void push_up(int p)
{
    
}

void push_down(int p)
{ 
    if (s[p].lazy != 0) {
        s[p<<1].lazy += s[p].lazy;
        s[p<<1|1].lazy += s[p].lazy;
        s[p].lazy = 0;
    }
}

void modify(int p, int l, int r, int val)
{
    if (l == s[p].l && r == s[p].r) {
        s[p].lazy += val;
    }
    else {
        int mid = (s[p].l + s[p].r) >> 1;
        push_down(p);
        if (r <= mid) {
            modify(p<<1, l, r, val);
        }
        else if (l > mid) {
            modify(p<<1|1, l, r, val);
        }
        else {
            modify(p<<1, l, mid, val);
            modify(p<<1|1, mid+1, r, val);
        }
        push_up(p);
    }
}

int query(int p)
{
    if (s[p].l == s[p].r) {
        return (bool)(s[p].lazy) * (H[s[p].r].y - H[s[p].l-1].y);
    }
    else {
        push_down(p);
        return query(p<<1) + query(p<<1|1);
    }
}

int main()
{
    int c, sum, ca = 0;
    while (scanf("%d %d", &N, &M), N|M) {
        for (int i = 1; i <= N; ++i) {
            scanf("%d %d %d %d", &e[i].x1, &e[i].y1, &e[i].x2, &e[i].y2); 
        }
        printf("Case %d:\n", ++ca);
        for (int t = 1; t <= M; ++t) {
            sum = 0; 
            mp.clear();
            scanf("%d", &Q);
            for (int j = 1, k = 0; j <= Q; ++j, k += 2) {
                scanf("%d", &c);
                Line[k].x = e[c].x1, Line[k+1].x = e[c].x2;
                Line[k].sign = 1, Line[k+1].sign = -1;
                // 定义 1为入边，-1为出边
                Line[k].y1 = e[c].y1, Line[k].y2 = e[c].y2;
                Line[k+1].y1 = e[c].y1, Line[k+1].y2 = e[c].y2;
                H[k].y = e[c].y1, H[k+1].y = e[c].y2;
            }
            sort(H, H+2*Q); // 对y轴坐标进行离散话
            cnt = unique(H, H+2*Q) - H; // 去重
            for (int i = 0; i < cnt; ++i) {
                mp[H[i].y] = i;
            }
            sort(Line, Line+2*Q);
            for (int i = 0; i < 2*Q; ++i) {
                Line[i].y1 = mp[Line[i].y1];
                Line[i].y2 = mp[Line[i].y2];
                // 将y坐标进行离散化
            }
            build(1, 0, cnt-1);  // 建立一个空的树来表示整个纵坐标的覆盖情况
            modify(1, Line[0].y1+1, Line[0].y2, Line[0].sign); 
            for (int i = 1; i < (Q << 1); ++i) { // 遍历每一个
                sum += query(1) * (Line[i].x - Line[i-1].x); 
                modify(1, Line[i].y1+1, Line[i].y2, Line[i].sign); 
            }
            printf("Query %d: %d\n", t, sum);
        }
        puts("");
    }
    return 0;
}