HDU 3265 Posters

pid=3265" target="_blank" style="">题目链接

题意:给定一些矩形海报。中间有孔。求贴海报的之后的海报覆盖面积并

思路:海报一张能够分割成4个矩形。然后就是普通的矩形面积并了,利用线段树维护就可以

代码:

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std; typedef long long ll; const int N = 50005; struct Node {
int l, r, len, cover;
int size() {return r - l + 1;}
} node[N * 4]; struct Line {
int l, r, y, flag;
Line() {}
Line(int l, int r, int y, int flag) {
this->l = l; this->r = r;
this->y = y; this->flag = flag;
}
} line[N * 8]; struct Rec {
int x1, y1, x2, y2;
Rec() {}
Rec(int x1, int y1, int x2, int y2) {
this->x1 = x1; this->y1 = y1;
this->x2 = x2; this->y2 = y2;
}
} rec[N * 4]; bool cmp(Line a, Line b) {
return a.y < b.y;
} int n;
int x[4], y[4]; #define lson(x) ((x<<1)+1)
#define rson(x) ((x<<1)+2) void pushup(int x) {
if (node[x].cover) node[x].len = node[x].size();
else if (node[x].l == node[x].r) node[x].len = 0;
else node[x].len = node[lson(x)].len + node[rson(x)].len;
} void build(int l, int r, int x = 0) {
node[x].l = l; node[x].r = r;
if (l == r) {
node[x].cover = node[x].len = 0;
return;
}
int mid = (l + r) / 2;
build(l, mid, lson(x));
build(mid + 1, r, rson(x));
pushup(x);
} void add(int l, int r, int v, int x = 0) {
if (l > r) return;
if (node[x].l >= l && node[x].r <= r) {
node[x].cover += v;
pushup(x);
return;
}
int mid = (node[x].l + node[x].r) / 2;
if (l <= mid) add(l, r, v, lson(x));
if (r > mid) add(l, r, v, rson(x));
pushup(x);
} int main() {
while (~scanf("%d", &n) && n) {
build(0, 50000);
int rn = 0, ln = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < 4; j++)
scanf("%d%d", &x[j], &y[j]);
rec[rn++] = Rec(x[0], y[0], x[1], y[2]);
rec[rn++] = Rec(x[0], y[2], x[2], y[3]);
rec[rn++] = Rec(x[0], y[3], x[1], y[1]);
rec[rn++] = Rec(x[3], y[2], x[1], y[3]);
}
for (int i = 0; i < rn; i++) {
line[ln++] = Line(rec[i].x1, rec[i].x2, rec[i].y1, 1);
line[ln++] = Line(rec[i].x1, rec[i].x2, rec[i].y2, -1);
}
n = ln;
sort(line, line + n, cmp);
ll ans = 0;
for (int i = 0; i < n; i++) {
if (i) ans += (ll)node[0].len * (line[i].y - line[i - 1].y);
add(line[i].l, line[i].r - 1, line[i].flag);
}
printf("%lld\n", ans);
}
return 0;
}

HDU 3265 Posters(线段树)的更多相关文章

  1. HDU 3265 Posters (线段树+扫描线)(面积并)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3265 给你n个中间被挖空了一个矩形的中空矩形,让你求他们的面积并. 其实一个中空矩形可以分成4个小的矩 ...

  2. poj_2528Mayor's posters(线段树)

    poj_2528Mayor's posters(线段树) 标签: 线段树 题目连接 Mayor's posters Time Limit: 1000MS Memory Limit: 65536K To ...

  3. hdu 4031 attack 线段树区间更新

    Attack Time Limit: 5000/3000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)Total Subm ...

  4. hdu 4288 离线线段树+间隔求和

    Coder Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Su ...

  5. hdu 3016 dp+线段树

    Man Down Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total S ...

  6. POJ.2528 Mayor's posters (线段树 区间更新 区间查询 离散化)

    POJ.2528 Mayor's posters (线段树 区间更新 区间查询 离散化) 题意分析 贴海报,新的海报能覆盖在旧的海报上面,最后贴完了,求问能看见几张海报. 最多有10000张海报,海报 ...

  7. HDU 3265 Posters ——(线段树+扫描线)

    第一次做扫描线,然后使我对线段树的理解发生了动摇= =..这个pushup写的有点神奇.代码如下: #include <stdio.h> #include <algorithm> ...

  8. hdu 1828 Picture(线段树 || 普通hash标记)

    http://acm.hdu.edu.cn/showproblem.php?pid=1828 Picture Time Limit: 6000/2000 MS (Java/Others)    Mem ...

  9. (中等) HDU 3265 Posters , 扫描线。

    Problem Description Ted has a new house with a huge window. In this big summer, Ted decides to decor ...

随机推荐

  1. sublime text3中的常用插件

    1.All Autocomplete Sublime Text 默认的 Autocomplete 功能只考虑当前的文件,而 AllAutocomplete 插件会搜索所有打开的文件来寻找匹配的提示词. ...

  2. How to recover after deleting the symbolic link libc.so.6?

    参考资料: http://stackoverflow.com/questions/12249547/how-to-recover-after-deleting-the-symbolic-link-li ...

  3. jstl经典用法

    jstl的forEach使用和set变量实现自增: <body> <c:set var="index" value="0" /> < ...

  4. c - 每位数字尾部加空格

    /* input:一个4位整数. output:每位整数后紧跟一个空格的字符串. */ char * insert(char *s) { int len = strlen(s); * len + ); ...

  5. 常用webservice网址

    http://www.gpsso.com/Main/ServiceList.aspx http://developer.51cto.com/art/200908/147125.htm 这里记录了几个常 ...

  6. NSURLSessionDataTask

    #import "ViewController.h" @interface ViewController ()<NSURLSessionDelegate,NSURLSessi ...

  7. jQuery操作元素

    通常,我们在创建元素时,会使用以下代码: var p = document.createElement("p"); p.innerText = "this is para ...

  8. uva 10167 - Birthday Cake

    题解:由于解太多,随机抓 A.B, 只要有符合就行了: (首先,Ax+By=0必须表示直线,即A.B不能同时为0:另外,要注意到直线不能过输入中的2N个点:检测点在直线的哪一侧,只需要简单的线性规划的 ...

  9. h5connect.js 腾讯云视频点播使用指南

    http://video.qcloud.com/download/docs/QVOD_Player_Web_SDK_Developer_Guide.pdf 腾讯云视频点播服务 Web播放器SDK开发指 ...

  10. n阶行列式计算----c语言实现(完结)

    花了半天时间,写了这个n阶行列式计算的程序,应该算是比较优美吧,有很多地方多次做了优化,程序占用内存不是很大,要是说小吧,也不合适,因为里边有一个递归,而且递归的深度还比较深.时间复杂度具体没有细看, ...