Picture（线段树 + 扫描线 + 离散化）

Picture

Time Limit: 2000MS		Memory Limit: 10000K
Total Submissions: 10229		Accepted: 5421

Description

A number of rectangular posters, photographs and other pictures of the same shape are pasted on a wall. Their sides are all vertical or horizontal. Each rectangle can be partially or totally covered by the others. The length of the boundary of the union of all rectangles is called the perimeter. 

Write a program to calculate the perimeter. An example with 7 rectangles is shown in Figure 1. 

The corresponding boundary is the whole set of line segments drawn in Figure 2. 

The vertices of all rectangles have integer coordinates. 

Input

Your program is to read from standard input. The first line contains the number of rectangles pasted on the wall. In each of the subsequent lines, one can find the integer coordinates of the lower left vertex and the upper right vertex of each rectangle. The values of those coordinates are given as ordered pairs consisting of an x-coordinate followed by a y-coordinate. 

0 <= number of rectangles < 5000 
All coordinates are in the range [-10000,10000] and any existing rectangle has a positive area.

Output

Your program is to write to standard output. The output must contain a single line with a non-negative integer which corresponds to the perimeter for the input rectangles.

Sample Input

7
-15 0 5 10
-5 8 20 25
15 -4 24 14
0 -6 16 4
2 15 10 22
30 10 36 20
34 0 40 16

Sample Output

228

 

      题意：

      给出 N（0 ~ 5000）个矩形，后给出每个矩形的左下坐标和右上坐标。所有的矩形给出之后，会有一个覆盖的轮廓，求出这个轮廓的周长。

 

      思路：

      线段树 + 扫描线 + 离散化。lc 维护左端点是否被覆盖，rc 维护右端点是否被覆盖，len 维护覆盖的长度，num 代表一共覆盖的顶点数， cover 代表覆盖区间的次数。将一个矩形的左竖边标记为 1，右竖边标记为 -1，注意算横边的时候，是用原来的 num 覆盖顶点数来算。之后再插入新的纵边，再查询总长 len 后，res +=（ len - 上一次的 len）。注意更新 len 的时候，是更新离散化前的长度值。标记要用 += 值来标记，而不能直接用 1 或者 -1 来替换区间，因为可能会有一个区间被覆盖了两次的可能，比如样例：

       2

       5 1 10 2

       6 1 15 2

       如果直接用 1 或者 -1 覆盖得出的答案是 12，若用 += 的话则才是正确答案 22。

 

      观察 push_up 函数：

void push_up (int node, int l, int r) {
        if (cover[node]) {
                len[node] = yy[r] - yy[l];
                lc[node] = rc[node] = 1;
                num[node] = 2; 
                //代表这个区间被覆盖
        } else if (r - l == 1) {
                len[node] = lc[node] =
                rc[node] = num[node] = 0;  
                //这个区间不被覆盖，且又是叶子节点
        } else {
                len[node] = len[node << 1] + len[node << 1 | 1];
                lc[node] = lc[node << 1];
                rc[node] = rc[node << 1 | 1];
                num[node] = num[node << 1] + num[node << 1 | 1];
                if (rc[node << 1] && lc[node << 1 | 1]) num[node] -= 2;
                //这个区间不被覆盖，且不是叶子节点
        }
}

 

     AC：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

const int MAX = 20005;

typedef struct {
        int x, y1, y2, temp;
} node;

node line[5005 * 2];
int yy[MAX];

int lc[MAX * 3], rc[MAX * 3], cover[MAX * 3];
int len[MAX * 3], num[MAX * 3];

bool cmp (node a, node b) {
        if (a.x != b.x) return a.x < b.x;
        return a.temp > b.temp;
}

void push_up (int node, int l, int r) {
        if (cover[node]) {
                len[node] = yy[r] - yy[l];
                lc[node] = rc[node] = 1;
                num[node] = 2;
        } else if (r - l == 1) {
                len[node] = lc[node] =
                rc[node] = num[node] = 0;
        } else {
                len[node] = len[node << 1] + len[node << 1 | 1];
                lc[node] = lc[node << 1];
                rc[node] = rc[node << 1 | 1];
                num[node] = num[node << 1] + num[node << 1 | 1];
                if (rc[node << 1] && lc[node << 1 | 1]) num[node] -= 2;
        }
}

void build (int node, int l, int r) {

        if (r - l == 1) {
                cover[node] = 0;
                lc[node] = rc[node] =
                len[node] = num[node] = 0;
        } else {
                int mid = (r + l) >> 1;
                build(node << 1, l, mid);
                build(node << 1 | 1, mid, r);
                push_up(node, l, r);
                cover[node] = 0;
        }
}

void updata (int node, int l, int r, int cl, int cr, int c) {

        if (cl > r || cr < l) return;
        if (cl <= l && cr >= r) {
                cover[node] += c;
                push_up(node, l, r);
                return;
        }
        if (r - l == 1) return;

        int mid = (r + l) >> 1;
        updata(node << 1, l, mid, cl, cr, c);
        updata(node << 1 | 1, mid, r, cl, cr, c);
        push_up(node, l, r);
}

int main () {
        int t;

        while (~scanf("%d", &t)) {

                int n = 0, ans = 0;
                while (t--) {
                        int x1, y1, x2, y2;
                        scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
                        line[n].x = x1;
                        line[n].y1 = y1;
                        line[n].y2 = y2;
                        line[n++].temp = 1;

                        line[n].x = x2;
                        line[n].y1 = y1;
                        line[n].y2 = y2;
                        line[n++].temp = -1;

                        yy[ans++] = y1;
                        yy[ans++] = y2;
                }

                sort(line, line + n, cmp);

                sort(yy, yy + ans);
                ans = unique(yy, yy + ans) - yy;

                build(1, 0, ans - 1);
                int res = 0, last = 0;
                for (int i = 0; i < n; ++i) {
                        int cl = lower_bound(yy, yy + ans, line[i].y1) - yy;
                        int cr = lower_bound(yy, yy + ans, line[i].y2) - yy;

                        if(i) res += num[1] * (line[i].x - line[i - 1].x);

                        updata(1, 0, ans - 1, cl, cr, line[i].temp);
                        res += abs(len[1] - last);

                        last = len[1];

                }

                printf("%d\n", res);
        }

        return 0;
}