线段树扫描线（矩阵周长并）——HDU 1828

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我是靠谱客的博主想人陪鸭子，最近开发中收集的这篇文章主要介绍线段树扫描线（矩阵周长并）——HDU 1828，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

对应HDU题目：点击打开链接

Picture

Time Limit: 2000MS

Memory Limit: 32768KB

64bit IO Format: %I64d & %I64u

Submit Status

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.

Please process to the end of file.

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

Source

题意：矩阵周长并

思路：线段树扫描线，先从下往上扫，求出所有横边；再从左往右扫，求出所有竖边，最后相加。还有别的办法有空补上。。。

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<algorithm>
#include<cstring>
#include<string>
#include<iostream>
const int MAXN=10000+10;
#define ms(x,y) memset(x,y,sizeof(x))
using namespace std;
int x[MAXN];
int y[MAXN];
int sum[MAXN<<2];
int cnt[MAXN<<2];
struct Line
{
int l,a,b;
int flag;
}line_y[MAXN],line_x[MAXN];
bool cmp(Line l1, Line l2)
{
return l1.l < l2.l;
}
void up(int rt, int left, int right, int *A)
{
if(cnt[rt]) sum[rt] = A[right+1] - A[left];
else if(left == right) sum[rt] = 0;
else sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}
void update(int rt, int left, int right, int l, int r, int flag, int *A)
{
if(l == left && right == r){
cnt[rt]+=flag;
up(rt, left, right, A);
return;
}
int mid=(left + right)>>1;
if(mid >= r) update(rt<<1, left, mid, l, r, flag, A);
else if(mid < l) update(rt<<1|1, mid+1, right, l, r, flag, A);
else{
update(rt<<1, left, mid, l, mid, flag, A);
update(rt<<1|1, mid+1, right, mid+1, r, flag, A);
}
up(rt, left, right, A);
}
int binary(int key, int n, int *a)
{
int l=0, r=n-1;
while(l<=r)
{
int mid=(l+r)>>1;
if(a[mid] == key) return mid;
if(a[mid] > key) r = mid - 1;
else l = mid + 1;
}
return -1;
}
int main()
{
//freopen("in.txt","r",stdin);
int n,w=0;
while(~scanf("%d", &n))
{
ms(x,0);
ms(y,0);
ms(cnt,0);
ms(sum,0);
int u=0;
int uu=0;
int n_x=0;
int n_y=0;
for(int i=0; i<n; i++){
int x1,y1,x2,y2;
scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
x[n_x++]=x1;
x[n_x++]=x2;
line_y[u].l=y1;
line_y[u].flag=1;
line_y[u].a=x1;
line_y[u++].b=x2;
line_y[u].l=y2;
line_y[u].flag=-1;
line_y[u].a=x1;
line_y[u++].b=x2;
y[n_y++]=y1;
y[n_y++]=y2;
line_x[uu].l=x1;
line_x[uu].flag=1;
line_x[uu].a=y1;
line_x[uu++].b=y2;
line_x[uu].l=x2;
line_x[uu].flag=-1;
line_x[uu].a=y1;
line_x[uu++].b=y2;
}
sort(x, x+n_x);
sort(y, y+n_y);
sort(line_y, line_y+u, cmp);
sort(line_x, line_x+uu, cmp);
int k=1;
for(int i=1; i<n_x; i++) if(x[i] != x[i-1]) x[k++] = x[i];
int kk=1;
for(int i=1; i<n_y; i++) if(y[i] != y[i-1]) y[kk++] = y[i];
int ans_y=0;
int last_y=0;
for(int i=0; i<u; i++){
int l=binary(line_y[i].a, k, x);
int r=binary(line_y[i].b, k, x);
update(1, 0, k-2, l, r-1, line_y[i].flag, x);
if(line_y[i].flag == 1){
ans_y += sum[1] - last_y;
}
else {
ans_y += abs(sum[1] - last_y);
}
last_y = sum[1];
}
ms(cnt,0);
ms(sum,0);
int ans_x=0;
int last_x=0;
for(int i=0; i<uu; i++){
int l=binary(line_x[i].a, kk, y);
int r=binary(line_x[i].b, kk, y);
update(1, 0, kk-2, l, r-1, line_x[i].flag, y);
if(line_x[i].flag == 1){
ans_x += sum[1] - last_x;
}
else {
ans_x += abs(sum[1] - last_x);
}
last_x = sum[1];
}
printf("%dn", ans_x + ans_y);
}
return 0;
}