POJ 2777 Count Color(线段树+二进制位运算)
来源:互联网 发布:windows什么系统好用 编辑:程序博客网 时间:2024/06/06 17:22
Count Color
Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 43938 Accepted: 13310
Description
Chosen Problem Solving and Program design as an optional course, you are required to solve all kinds of problems. Here, we get a new problem.
There is a very long board with length L centimeter, L is a positive integer, so we can evenly divide the board into L segments, and they are labeled by 1, 2, ... L from left to right, each is 1 centimeter long. Now we have to color the board - one segment with only one color. We can do following two operations on the board:
1. "C A B C" Color the board from segment A to segment B with color C.
2. "P A B" Output the number of different colors painted between segment A and segment B (including).
In our daily life, we have very few words to describe a color (red, green, blue, yellow…), so you may assume that the total number of different colors T is very small. To make it simple, we express the names of colors as color 1, color 2, ... color T. At the beginning, the board was painted in color 1. Now the rest of problem is left to your.
There is a very long board with length L centimeter, L is a positive integer, so we can evenly divide the board into L segments, and they are labeled by 1, 2, ... L from left to right, each is 1 centimeter long. Now we have to color the board - one segment with only one color. We can do following two operations on the board:
1. "C A B C" Color the board from segment A to segment B with color C.
2. "P A B" Output the number of different colors painted between segment A and segment B (including).
In our daily life, we have very few words to describe a color (red, green, blue, yellow…), so you may assume that the total number of different colors T is very small. To make it simple, we express the names of colors as color 1, color 2, ... color T. At the beginning, the board was painted in color 1. Now the rest of problem is left to your.
Input
First line of input contains L (1 <= L <= 100000), T (1 <= T <= 30) and O (1 <= O <= 100000). Here O denotes the number of operations. Following O lines, each contains "C A B C" or "P A B" (here A, B, C are integers, and A may be larger than B) as an operation defined previously.
Output
Ouput results of the output operation in order, each line contains a number.
Sample Input
2 2 4C 1 1 2P 1 2C 2 2 2P 1 2
Sample Output
21
题意:
给一个固定长度为L的画板
有两个操作:
C A B C:区间AB内涂上颜色C。
P A B:查询区间AB内颜色种类数。
思路:
首先显然是要线段树。
每个节点有如下参数:
l,r 表示区间。
int color表示颜色,对于颜色要用位运算。
AC代码:
#include <stdio.h>#include <string.h>#include <algorithm>#include <iostream>using namespace std;const int M=112345;struct node{ int l,r; int c; bool flag; //标记此区间是否为一种颜色覆盖}p[M<<2];void pushup(int root)//位运算求区间颜色种类{ p[root].c=p[root<<1].c|p[root<<1|1].c;}void build(int root,int b,int e){ p[root].l=b; p[root].r=e; p[root].c=1; p[root].flag=1; if(b==e)return ; int mid=(b+e)/2; build(root<<1,b,mid); build(root<<1|1,mid+1,e); pushup(root);}void pushdown(int root){ p[root<<1].c=p[root].c; p[root<<1].flag=1; p[root<<1|1].c=p[root].c; p[root<<1|1].flag=1; p[root].flag=0;}void updata(int root,int b,int e,int c){ if(p[root].l>=b&&p[root].r<=e) { p[root].c=c; p[root].flag=1; return ; } if(p[root].c==c)return ;//剪枝 if(p[root].flag)pushdown(root); int mid=(p[root].l+p[root].r)/2; if(mid>=e) updata(root<<1,b,e,c); else if(mid<b) updata(root<<1|1,b,e,c); else { updata(root<<1,b,mid,c); updata(root<<1|1,mid+1,e,c); } pushup(root);}int sum;void query(int root,int b,int e){ if(p[root].l>=b&&p[root].r<=e) { sum |= p[root].c; return ; } if(p[root].flag) { sum|=p[root].c; return ; } int mid=(p[root].l+p[root].r)/2; if(mid>=e) query(root<<1,b,e); else if(mid<b) query(root<<1|1,b,e); else { query(root<<1,b,mid); query(root<<1|1,mid+1,e); }}int solve(){ int ans=0; while(sum) //求sum中1的个数 { if(sum&1) ans++; sum>>=1; } return ans;}void swap(int &a,int &b){ int t; t=a; a=b; b=t;}int main(){ int n,m,q; while(~scanf("%d%d%d",&n,&m,&q)) { build(1,1,n); while(q--) { char c[2]; scanf("%s",c); if(c[0]=='C') { int a,b,c; scanf("%d%d%d",&a,&b,&c); if(a>b)swap(a,b); updata(1,a,b,1<<(c-1)); } else { sum=0; int a,b; scanf("%d%d",&a,&b); query(1,a,b); // printf("%d\n",sum); printf("%d\n",solve()); } } }}
0 0
- POJ 2777 Count Color(线段树+二进制位运算)
- POJ 2777 Count Color (线段树+位运算)
- POJ 2777 Count Color(线段树+位运算)
- POJ 2777 Count Color(线段树+位运算)
- POJ 2777 Count Color 线段树无位运算
- poj 2777 -- Count Color ( 线段树 )
- poj - 2777 - Count Color(线段树)
- POJ 2777 Count Color(线段树)
- poj 2777 Count Color(线段树)
- POJ 2777--Count Color(线段树)
- POJ 2777 Count Color(线段树)
- POJ 2777 Count Color(线段树 )
- POJ 2777 Count Color(线段树)
- POJ 2777 Count Color (线段树)
- [POJ 2777]Count Color(线段树)
- POJ 2777 Count Color(线段树)
- POJ 2777-Count Color-(线段树)
- POJ 2777 Count Color(线段树+位运算优化)
- 《21天实战Caffe》第六章习题6.4 自己的手写体数字图片送入lenet测试
- 1020 Tree Traversals --二叉树乱搞
- 算法 最优分解问题
- 8.7 泛型类型的继承规则
- C#编程学习笔记1
- POJ 2777 Count Color(线段树+二进制位运算)
- HDU-2097 Sky数
- 8.8 通配符类型
- 在服务器GPU73上编译安装Caffe
- 8.9 反射和泛型
- 【Linux命令行与shell脚本编程】教程三——切换目录
- 参数服务器——分布式机器学习的新杀器
- org.apache.ibatis.exceptions.TooManyResultsException: Expected one result (or null) to be returned b
- Introduction to Programming with c++ 13-7 BinaryIO