POJ 2777 Count Color(线段树+位运算)
题目链接:http://poj.org/problem?id=2777
Description
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
defined previously.
Output
Sample Input
2 2 4
C 1 1 2
P 1 2
C 2 2 2
P 1 2
Sample Output
2
1
Source
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A--B内涂上颜色C。
P A B:查询区间AB内颜色种类数。
PS:
此题和HDU:5023是类似的!
附题解:http://blog.csdn.net/u012860063/article/details/39434665
代码例如以下:
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std; #define lson l , mid , rt << 1
#define rson mid + 1 , r , rt << 1 | 1
#define LL int const int maxn = 110017;
LL add[maxn<<2];
LL sum[maxn<<2];
void PushUp(int rt)
{
//把当前结点的信息更新到父结点
sum[rt] = sum[rt<<1] | sum[rt<<1|1];//总共的颜色
}
void PushDown(int rt,int m)
{
if(add[rt])
{
add[rt<<1] = add[rt];
add[rt<<1|1] = add[rt];
sum[rt<<1] = add[rt];
sum[rt<<1|1] = add[rt];
add[rt] = 0;//将标记向儿子节点移动后,父节点的延迟标记去掉
//传递后,当前节点标记域清空
}
}
void build(int l,int r,int rt)
{
add[rt] = 0;//初始化为全部结点未被标记
if (l == r)
{
sum[rt] = 1;//初始颜色为1
return ;
}
int mid = (l + r) >> 1;
build(lson);
build(rson);
PushUp(rt);
}
void update(int L,int R,int c,int l,int r,int rt)
{
if (L <= l && r <= R)
{
add[rt] =1<<(c-1);//位运算左移表示有某种颜色
sum[rt] =1<<(c-1);
return ;
}
PushDown(rt , r - l + 1);//----延迟标志域向下传递
int mid = (l + r) >> 1;
if (L <= mid)
update(L , R , c , lson);//更新左儿子
if (mid < R)
update(L , R , c , rson);//更新右儿子
PushUp(rt);
}
LL query(int L,int R,int l,int r,int rt)
{
if (L <= l && r <= R)
{
return sum[rt];
}
//要取rt子节点的值时,也要先把rt的延迟标记向下移动
PushDown(rt , r - l + 1);
int mid = (l + r) >> 1;
LL ret = 0;
if (L <= mid)
ret |= query(L , R , lson);
if (mid < R)
ret |= query(L , R , rson);
return ret;
}
int main()
{
int L, T, O;
int a, b, c;
while(~scanf("%d%d%d",&L,&T,&O))
{
build(1, L, 1);//建树
while(O--)//Q为询问次数
{
char op[2];
scanf("%s",op);
if(op[0] == 'P')
{
scanf("%d%d",&a,&b);
if(a > b)
{
int t = a;
a = b;
b = t;
}
LL tt=query(a, b, 1, L, 1);
int ans = 0;
while(tt)
{
if(tt&1)
{
ans++;
}
tt>>=1;
}
printf("%d\n",ans);
}
else
{
scanf("%d%d%d",&a,&b,&c);
if(a > b)
{
int t = a;
a = b;
b = t;
}
update(a, b, c, 1, L, 1);
}
}
}
return 0;
}
POJ 2777 Count Color(线段树+位运算)的更多相关文章
- poj 2777 Count Color - 线段树 - 位运算优化
Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 42472 Accepted: 12850 Description Cho ...
- poj 2777 Count Color(线段树区区+染色问题)
题目链接: poj 2777 Count Color 题目大意: 给出一块长度为n的板,区间范围[1,n],和m种染料 k次操作,C a b c 把区间[a,b]涂为c色,P a b 查 ...
- poj 2777 Count Color(线段树)
题目地址:http://poj.org/problem?id=2777 Count Color Time Limit: 1000MS Memory Limit: 65536K Total Subm ...
- poj 2777 Count Color(线段树、状态压缩、位运算)
Count Color Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 38921 Accepted: 11696 Des ...
- Count Color(线段树+位运算 POJ2777)
Count Color Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 39917 Accepted: 12037 Descrip ...
- POJ 2777 Count Color(线段树之成段更新)
Count Color Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 33311 Accepted: 10058 Descrip ...
- POJ 2777 Count Color (线段树成段更新+二进制思维)
题目链接:http://poj.org/problem?id=2777 题意是有L个单位长的画板,T种颜色,O个操作.画板初始化为颜色1.操作C讲l到r单位之间的颜色变为c,操作P查询l到r单位之间的 ...
- [poj2777] Count Color (线段树 + 位运算) (水题)
发现自己越来越傻逼了.一道傻逼题搞了一晚上一直超时,凭啥子就我不能过??? 然后发现cin没关stdio同步... Description Chosen Problem Solving and Pro ...
- POJ P2777 Count Color——线段树状态压缩
Description Chosen Problem Solving and Program design as an optional course, you are required to sol ...
- POJ 2777 Count Color(段树)
职务地址:id=2777">POJ 2777 我去.. 延迟标记写错了.标记到了叶子节点上.. . . 这根本就没延迟嘛.. .怪不得一直TLE... 这题就是利用二进制来标记颜色的种 ...
随机推荐
- poj 2417 Discrete Logging(A^x=B(mod c),普通baby_step)
http://poj.org/problem?id=2417 A^x = B(mod C),已知A,B.C.求x. 这里C是素数,能够用普通的baby_step. 在寻找最小的x的过程中,将x设为i* ...
- 搜索引擎排名不友好的五个地点-SEO
搜索引擎(百度/谷歌/雅虎)排名不友好的五个地点 别的站点,推断标准和考核得分点是不 一样的,避免对百度排名不友好的五种站点操作 你的站点是否在这五种站点里,决定你的站点能否获得排 名. 1.有没有同 ...
- python字典构造函数dict(mapping)解析
Python字典的构造函数有三个,dict().dict(**args).dict(mapping),当中第一个.第二个构造函数比較好理解也比較easy使用, 而dict(mapping)这个构造函数 ...
- java通过抛异常来返回提示信息
结论:如果把通过抛异常的方式得到提示信息,可以使用java.lang.Throwable中的构造函数: public Throwable(String message) { fillInStackTr ...
- crm创建报告补充导航
报告导航实现动态交互体验报告. 通过使用各种类型的操作的,报告允许用户导航到特定的报告.Microsoft Dynamics CRM 记录或其它网站 动态钻取到 Microsoft Dynamics ...
- hdu 4115 石头剪子布(2-sat问题)
/* 意甲冠军:石头剪子布,目前已知n周围bob会有什么,对alice限制.供u,v,w:设w=0说明a,b回合必须出的一样 否则,必须不一样.alice假设输一回合就输了,否则就赢了 解: 2-sa ...
- Cocos2d-x3.0 RenderTexture(一) 保存
.h #include "cocos2d.h" #include "cocos-ext.h" #include "ui/CocosGUI.h" ...
- Git 常用命令手记 及 Github协同流程(转)
符号约定俗成:<xxx> 自定义内容xxx:[xxx] xxx为可选项:[<xxx>] 自定义内容xxx且为可选项. 说明/备注 命令 备注 保存更新 git add [-i] ...
- POJ 3692:Kindergarten(最大的使命)
id=3692">Kindergarten Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 4920 Ac ...
- 划分数 (DP)
输入: n=4 m=3 M=10000 输出: 4 (1+1+2=1+3=2+2=4) 复杂度(nm) int n,m; int a[MAX]; int dp[MAX][MAX]; //数组 void ...