线段树区间更新——POJ 2777

对应POJ题目：点击打开链接

Count Color

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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

题意：1~N开始被刷上颜色1，有O个操作，C  a,b,c 表示把a~b区间刷成颜色c;   P a,b 表示求a~b区间共有多少种颜色。

思路：线段树；color[rt] 表示结点rt有哪几种颜色，用二进制表示；查询时不断并上符合要求的区间的值就可以了。这道题学习了求一个整数有多少个1的几种方法，还是有点收获的。。。

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<algorithm>
#include<cstring>
#include<string>
#include<iostream>
#define ms(x,y) memset(x,y,sizeof(x))
#define N 100000+10
const int MAXN=1000+10;
const int INF=1<<30;
using namespace std;
int color[N<<2];
int ans;

#if 0

int cal(int x)//求一个整数中1的个数普通方法,速度一般
{
	int cnt=0;
	while(x)
	{
		if(x&1) cnt++;
		x>>=1;
	}
	return cnt;
}

/*
一个整数不为0，那么这个整数至少有一位是1。如果我们把这个整数减去1，那么原来处在整数最右边的1就会变成0，原来在1后面的所有 的0都会变成1。其余的所有位将不受到影响。举个例子：一个二进制数1100，从右边数起的第三位是处于最右边的一个1。减去1后，第三位变成0，它后面 的两位0变成1，而前面的1保持不变，因此得到结果是1011。
我们发现减1的结果是把从最右边一个1开始的所有位都取反了。这个时候如果我们再把原来的整数和减去1之后的结果做与运算，从原来整数最右边一个1那一位 开始所有位都会变成0。如1100&1011=1000。也就是说，把一个整数减去1，再和原整数做与运算，会把该整数最右边一个1变成0。那么 一个整数的二进制有多少个1，就可以进行多少次这样的操作。
 */
int cal(int x)//速度较快
{
	int cnt=0;
	while(x)
	{
		++cnt;
		x=(x-1)&x;
	}
	return cnt;
}

#endif

//HAKMEM算法：
int cal(unsigned x)//速度更快
{
    unsigned n;
    n = (x >> 1) & 033333333333;
    x = x - n;
    n = (n >> 1) & 033333333333;
    x = x - n;
    x = (x + (x >> 3)) & 030707070707;
    x = x%63;
    return x;
} 

void updata(int rt, int left, int right, int l, int r, int c)
{
	if(left==l && right==r){
		color[rt]=1<<(c-1);//这里开始是直接=c,好久才发现问题。。。
		return;
	}
	int cnt=cal(color[rt]);//如果该结点只有一种颜色，则向下更新
	if(cnt==1){
		color[rt<<1]=color[rt];
		color[rt<<1|1]=color[rt];
	}
	int mid=(left+right)>>1;
	if(mid>=r) updata(rt<<1, left, mid, l, r, c);
	else if(mid<l) updata(rt<<1|1, mid+1, right, l, r, c);
	else{
		updata(rt<<1, left, mid, l, mid, c);
		updata(rt<<1|1, mid+1, right, mid+1, r, c);
	}
	color[rt]=color[rt<<1]|color[rt<<1|1];//合并
}

void query(int rt, int left, int right, int l, int r)
{
	if(l==left && right==r){
		ans|=color[rt];
		return;
	}
	if(left==right) return;
	int cnt=cal(color[rt]);
	if(cnt==1){
		color[rt<<1]=color[rt];
		color[rt<<1|1]=color[rt];
	}
	int mid=(left+right)>>1;
	if(mid>=r) query(rt<<1, left, mid, l, r);
	else if(mid<l) query(rt<<1|1, mid+1, right, l, r);
	else{
		query(rt<<1, left, mid, l, mid);
		query(rt<<1|1, mid+1, right, mid+1, r);
	}
}

int main()
{
	//freopen("in.txt","r",stdin);
	int L,T,O;
	int a,b,c;
	char op;
	scanf("%d%d%d", &L,&T,&O);
	for(int i=1; i<(N<<2); i++) color[i]=1;
	while(O--)
	{
		scanf("%s", &op);
		if(op=='C'){
			scanf("%d%d%d", &a,&b,&c);
			if(b<a){//不断WA，坑！网上看人的才知道a可能大于b。。。
				int tmp=a; a=b; tmp=a;
			}
			updata(1,1,N,a,b,c);
		}
		else{
			scanf("%d%d", &a,&b);
			if(b<a){
				int tmp=a; a=b; tmp=a;
			}
			ans=0;
			query(1,1,N,a,b);
			printf("%d\n", cal(ans));
		}
	}
	return 0;
}