poj2777 Count Color

Count Color

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 40234		Accepted: 12141

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

Source

POJ Monthly--2006.03.26,dodo

首先这道题是区间颜色覆盖问题，很容易想到线段树。然后我们又发现颜色数≤30，所以这道题就可以和位运算集合到一起，这个思路很巧妙。

注意：A可能大于B，这时需要交换两个数。(我一开始没发现结果RE了......T T)

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#define F(i,j,n) for(int i=j;i<=n;i++)
#define D(i,j,n) for(int i=j;i>=n;i--)
#define LL long long
#define MAXN 100005
#define pa pair<int,int>
using namespace std;
int l,c,n,x,y,z;
char ch;
struct tree_type
{
	int l,r,w,tag;
}t[MAXN*4];
int read()
{
	int ret=0,flag=1;char ch=getchar();
	while (ch<'0'||ch>'9'){if (ch=='-') flag=-1;ch=getchar();}
	while (ch>='0'&&ch<='9'){ret=ret*10+ch-'0';ch=getchar();}
	return ret*flag;
}
void pushup(int k)
{
	t[k].w=t[k<<1].w|t[k<<1|1].w;
}
void update(int k,int z)
{
	t[k].tag=t[k].w=z;
}
void pushdown(int k)
{
	if (t[k].tag==0) return;
	update(k<<1,t[k].tag);
	update(k<<1|1,t[k].tag);
	t[k].tag=0;
}
void build(int k,int x,int y)
{
	t[k].l=x;
	t[k].r=y;
	t[k].tag=0;
	if (x==y) {t[k].w=1;return;}
	int mid=(x+y)>>1;
	build(k<<1,x,mid);
	build(k<<1|1,mid+1,y);
	pushup(k);
}
void cover(int k,int x,int y,int z)
{
	if (t[k].l==x&&t[k].r==y) {update(k,z);return;}
	pushdown(k);
	int mid=(t[k].l+t[k].r)>>1;
	if (mid>=y) cover(k<<1,x,y,z);
	else if (mid+1<=x) cover(k<<1|1,x,y,z);
	else cover(k<<1,x,mid,z),cover(k<<1|1,mid+1,y,z);
	pushup(k);
}
int query(int k,int x,int y)
{
	if (t[k].l==x&&t[k].r==y) return t[k].w;
	pushdown(k);
	int mid=(t[k].l+t[k].r)>>1;
	if (mid>=y) return query(k<<1,x,y);
	else if (mid+1<=x) return query(k<<1|1,x,y);
	else return (query(k<<1,x,mid)|query(k<<1|1,mid+1,y));
}
int calc(int x)
{
	int ans=0;
	for(;x;x>>=1) ans+=x&1;
	return ans;
}
int main()
{
	l=read();c=read();n=read();
	build(1,1,l);
	F(i,1,n)
	{
		ch=getchar();while (ch!='C'&&ch!='P') ch=getchar();
		x=read();y=read();
		if (x>y) swap(y,x);
		if (ch=='C') z=read(),cover(1,x,y,1<<(z-1));
		else printf("%d\n",calc(query(1,x,y)));
	}
	return 0;
}