POJ-2777Count Color 线段树+位移
这道题对于我这样的初学者还是有点难度的不过2遍A了还是很开心，下面说说想法……
Count Color
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 40302 Accepted: 12161
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.
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 4
C 1 1 2
P 1 2
C 2 2 2
P 1 2
Sample Output
2
1
题目大意：
往画板上染色，画板长L（1<=L<=100000）颜色共有T种（1<=T<=30）给你O条操作（1<=O<=100000）操作分两种
（1）“C A B C”操作C：将【A，B】染成C色（开始时，画板都是颜色1）
（2）“P A B”操作P：【A，B】范围内颜色种数
（此题有个蛋疼的地方，读入的AB可能顺序颠倒...也是略坑）
这道题用位移来做，如果染成t种颜色，就把颜色左移t-1位，位移运算在这道题里完美展现，所以以后还是得注重这种表示方法，了解多了，自然做题顺畅
用二进制表示对应的区间涂了第几种颜色，这样每个区间除了延迟标记外，可以再开一个数组统计当前涂了哪几种颜色。这样就和一般的线段树一样了。
（这种思想值得学习）
最后再统计一下1的数目
下面是代码：
//这波位移非常的nice啊！
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define maxl 100001
int color[maxl<<2]={0},sum[maxl<<2]={0};
void updata(int now)
{
sum[now]=sum[now<<1]|sum[now<<1|1];
}//or
void pushdown(int now)
{
if (color[now]!=0)
{
color[now<<1]=color[now<<1|1]=color[now];
sum[now<<1]=color[now];
sum[now<<1|1]=color[now];
color[now]=0;
}
}
void build(int l,int r,int now)
{
color[now]=1;
if (l==r) return;
int mid=(l+r)>>1;
build(l,mid,now<<1);
build(mid+1,r,now<<1|1);
updata(now);
}
void change(int L,int R,int l,int r,int now,int newcolor)
{
if (L<=l && R>=r)
{
color[now]=1<<(newcolor-1);
sum[now]=color[now];
return;
}//位移表示颜色
int mid=(l+r)>>1;
pushdown(now);
if (L<=mid)
change(L,R,l,mid,now<<1,newcolor);
if (R>mid)
change(L,R,mid+1,r,now<<1|1,newcolor);
updata(now);
}
int query(int L,int R,int l,int r,int now)
{
if (L<=l && R>=r)
return sum[now];
pushdown(now);
int mid=(l+r)>>1;
int ans=0;
if (L<=mid)
ans=ans|query(L,R,l,mid,now<<1);
if (R>mid)
ans=ans|query(L,R,mid+1,r,now<<1|1);//要对结果取or
return ans;
}
int main()
{
int l,t,o;
while(~scanf("%d%d%d",&l,&t,&o))
{
build(1,l,1);
for (int i=1; i<=o; i++)
{
char command[2];
int left,right,data;
scanf("%s",&command);
if (command[0]=='C')
{
scanf("%d%d%d",&left,&right,&data);
if (left>right)
{
int temp=left;
left=right;
right=temp;
}
change(left,right,1,l,1,data);
}
else
{
scanf("%d%d",&left,&right);
if (left>right)
{
int temp=left;
left=right;
right=temp;
}
int ans=query(left,right,1,l,1);
int answer=0;
while (ans>0)
{
if (ans & 1)
answer++;
ans=ans>>1;
}//这里表示颜色数
printf("%d\n",answer);
}
}
printf("\n");
}
return 0;
}
