题目链接:https://vjudge.net/problem/ZOJ-1610

Painting some colored segments on a line, some previously painted segments may be covered by some the subsequent ones.

Your task is counting the segments of different colors you can see at last.

Input

The first line of each data set contains exactly one integer n, 1 <= n <= 8000, equal to the number of colored segments.

Each of the following n lines consists of exactly 3 nonnegative integers separated by single spaces:

x1 x2 c

x1 and x2 indicate the left endpoint and right endpoint of the segment, c indicates the color of the segment.

All the numbers are in the range [0, 8000], and they are all integers.

Input may contain several data set, process to the end of file.

<b< dd="">

Output

Each line of the output should contain a color index that can be seen from the top, following the count of the segments of this color, they should be printed according to the color index.

If some color can't be seen, you shouldn't print it.

Print a blank line after every dataset.

<b< dd="">

Sample Input

5
0 4 4
0 3 1
3 4 2
0 2 2
0 2 3
4
0 1 1
3 4 1
1 3 2
1 3 1
6
0 1 0
1 2 1
2 3 1
1 2 0
2 3 0
1 2 1

<b< dd="">

Sample Output

1 1
2 1
3 1

1 1

0 2
1 1

题解:

问最终用多少段颜色相同的区域。经典的区间染色问题。

写法一:

 #include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
using namespace std;
typedef long long LL;
const double EPS = 1e-;
const int INF = 2e9;
const LL LNF = 2e18;
const int MAXN = 1e4+; int val[MAXN<<];
int num[MAXN], color[MAXN]; void push_down(int u)
{
if(val[u]>=)
{
val[u*] = val[u*+] = val[u];
val[u] = -;
}
} void set_val(int u, int l, int r, int x, int y, int _val)
{
if(x<=l && r<=y)
{
val[u] = _val;
return;
} push_down(u);
int mid = (l+r)>>;
if(x<=mid) set_val(u*, l, mid, x, y, _val);
if(y>=mid+) set_val(u*+, mid+, r, x, y, _val);
} void query(int u, int l, int r)
{
if(l==r)
{
color[l] = val[u];
return;
} push_down(u);
int mid = (l+r)>>;
query(u*, l, mid);
query(u*+, mid+, r);
} int main()
{
int m;
while(scanf("%d", &m)!=EOF)
{
memset(val, -, sizeof(val));
for(int i = ; i<=m; i++)
{
int x, y, z;
scanf("%d%d%d", &x, &y, &z);
if(x<y) set_val(, , , x+, y, z);
} query(, , );
memset(num, , sizeof(num));
for(int i = ; i<=; i++)
if(color[i]!=- && (i== || color[i]!=color[i-]))
num[color[i]]++; for(int i = ; i<=; i++)
if(num[i])
printf("%d %d\n", i, num[i]);
printf("\n");
}
}

写法二:

 #include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
using namespace std;
typedef long long LL;
const double EPS = 1e-;
const int INF = 2e9;
const LL LNF = 2e18;
const int MAXN = 1e4+; int val[MAXN<<];
int num[MAXN]; void push_down(int u)
{
if(val[u]>=)
{
val[u*] = val[u*+] = val[u];
val[u] = -;
}
} void set_val(int u, int l, int r, int x, int y, int _val)
{
if(x<=l && r<=y)
{
val[u] = _val;
return;
} push_down(u);
int mid = (l+r)>>;
if(x<=mid) set_val(u*, l, mid, x, y, _val);
if(y>=mid+) set_val(u*+, mid+, r, x, y, _val);
} int pre;
void query(int u, int l, int r)
{
if(l==r)
{
if(val[u]>= && val[u]!=pre)
num[val[u]]++;
pre = val[u];
return;
} push_down(u);
int mid = (l+r)>>;
query(u*, l, mid);
query(u*+, mid+, r);
} int main()
{
int m;
while(scanf("%d", &m)!=EOF)
{
memset(val, -, sizeof(val));
for(int i = ; i<=m; i++)
{
int x, y, z;
scanf("%d%d%d", &x, &y, &z);
if(x<y) set_val(, , , x+, y, z);
} memset(num, , sizeof(num));
pre = -;
query(, , );
for(int i = ; i<=; i++)
if(num[i])
printf("%d %d\n", i, num[i]);
printf("\n");
}
}

ZOJ1610 Count the Colors —— 线段树 区间染色的更多相关文章

  1. [ZOJ1610]Count the Colors(线段树,区间染色,单点查询)

    题目链接:http://www.icpc.moe/onlinejudge/showProblem.do?problemCode=1610 题意:给一个长8000的绳子,向上染色.一共有n段被染色,问染 ...

  2. ZOJ 1610 Count the Colors(线段树,区间覆盖,单点查询)

    Count the Colors Time Limit: 2 Seconds      Memory Limit: 65536 KB Painting some colored segments on ...

  3. zoj 1610 Count the Colors 线段树区间更新/暴力

    Count the Colors Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://acm.zju.edu.cn/onlinejudge/show ...

  4. ZOJ-1610 Count the Colors ( 线段树 )

    题目链接: http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1610 Description Painting some co ...

  5. ZOJ 1610.Count the Colors-线段树(区间染色、区间更新、单点查询)-有点小坑(染色片段)

    ZOJ Problem Set - 1610 Count the Colors Time Limit: 2 Seconds      Memory Limit: 65536 KB Painting s ...

  6. 线段树区间染色 ZOJ 1610

    Count the Colors ZOJ - 1610 传送门 线段树区间染色求染色的片段数 #include <cstdio> #include <iostream> #in ...

  7. 【POJ 2777】 Count Color(线段树区间更新与查询)

    [POJ 2777] Count Color(线段树区间更新与查询) Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 4094 ...

  8. HDU3974 Assign the task(多叉树转换为线段+线段树区间染色)

    题目大意:有n个人,给你他们的关系(老板和员工),没有直属上司的人就是整个公司的领导者,这意味着n个人形成一棵树(多叉树).当一个人被分配工作时他会让他的下属也做同样的工作(并且立即停止手头正在做的工 ...

  9. hdu 5023(线段树区间染色,统计区间内颜色个数)

    题目描述:区间染色问题,统计给定区间内有多少种颜色? 线段树模板的核心是对标记的处理 可以记下沿途经过的标记,到达目的节点之后一块算,也可以更新的时候直接更新到每一个节点 Lazy操作减少修改的次数( ...

随机推荐

  1. 利用virtualbox中的虚机制作主机启动盘

    制作镜像的过程: 第一步:1.Windows下先下载安装virtualbox usb3.0驱动:https://download.virtualbox.org/virtualbox/5.2.20/Or ...

  2. 【瞎扯】我的OI之路

    这里大概是一些我自己对我的OI之路的一些记录. 2015.11不知道哪一天-- 我听说了"编程". 当时还不懂得啥是信息学竞赛,以为这只是纯粹的程序设计.后来才明白信息学竞赛是算法 ...

  3. Leetcode 241.为运算表达式设计优先级

    为运算表达式设计优先级 给定一个含有数字和运算符的字符串,为表达式添加括号,改变其运算优先级以求出不同的结果.你需要给出所有可能的组合的结果.有效的运算符号包含 +, - 以及 * . 示例 1: 输 ...

  4. php引入PHPMailer发送邮件

    昨天做了一个发送邮件的功能,如果直接用mail()函数,需要拥有自己的邮件服务器,所有引入PHPMailer类方便快捷,简单写一下开发步骤: 一.拥有自己的邮箱账号(作为发件人邮箱) 分两种情况: 1 ...

  5. POJ 1509 循环同构的最小表示法

    题目大意: 给定一个字符串,可以把一段尾部接到头部,这样找到一个最小的字符串 方案一: 利用循环同构中找最小表示的方法来解决 论文参考http://wenku.baidu.com/view/438ca ...

  6. Python的另一种开发环境--Anaconda中的Spyder

    本文作者LucyGill,转载请注明出处(虽然我觉得并不会有人转载). 刚开始学Python的时候,我用的是其自带的idle(安装Python后,在开始菜单里可以找到),后来发现在eclipse中设置 ...

  7. 11-Js类和对象

    <!DOCTYPE html> <html> <head> <meta charset="UTF-8"> <title> ...

  8. Java的发送邮件

    以下内容引用自http://wiki.jikexueyuan.com/project/java/sending-email.html: 用Java应用程序来发送一封电子邮件是足够简单的,但是开始时应该 ...

  9. curl的使用(from 阮一峰)

    1.   http://www.ruanyifeng.com/blog/2011/09/curl.html 2.   https://curl.haxx.se/docs/httpscripting.h ...

  10. CentOS 7.0安装Zimbra 8.6邮件服务器

    Zimbra的核心产品是Zimbra协作套件(Zimbra Collaboration Suite,简称ZCS). 系统:Centos7 ip地址:192.168.127.131 安装前准备 1.关闭 ...