HDU 5023 A Corrupt Mayor's Performance Art

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最新推荐文章于 2020-08-04 17:30:37 发布

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分类专栏：线段树文章标签：区间更新

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本文解析了一道关于在线评测系统的编程题目，主要内容包括对题目要求的理解、数据结构的设计及算法实现思路。通过使用线段树数据结构进行区间更新和查询操作，解决了题目中的更新和查询颜色种类的问题。

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A Corrupt Mayor's Performance Art

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 100000/100000 K (Java/Others)
Total Submission(s): 3098 Accepted Submission(s): 1126

Problem Description

Corrupt governors always find ways to get dirty money. Paint something, then sell the worthless painting at a high price to someone who wants to bribe him/her on an auction, this seemed a safe way for mayor X to make money.

Because a lot of people praised mayor X's painting(of course, X was a mayor), mayor X believed more and more that he was a very talented painter. Soon mayor X was not satisfied with only making money. He wanted to be a famous painter. So he joined the local painting associates. Other painters had to elect him as the chairman of the associates. Then his painting sold at better price.

The local middle school from which mayor X graduated, wanted to beat mayor X's horse fart(In Chinese English, beating one's horse fart means flattering one hard). They built a wall, and invited mayor X to paint on it. Mayor X was very happy. But he really had no idea about what to paint because he could only paint very abstract paintings which nobody really understand. Mayor X's secretary suggested that he could make this thing not only a painting, but also a performance art work.

This was the secretary's idea:

The wall was divided into N segments and the width of each segment was one cun(cun is a Chinese length unit). All segments were numbered from 1 to N, from left to right. There were 30 kinds of colors mayor X could use to paint the wall. They named those colors as color 1, color 2 .... color 30. The wall's original color was color 2. Every time mayor X would paint some consecutive segments with a certain kind of color, and he did this for many times. Trying to make his performance art fancy, mayor X declared that at any moment, if someone asked how many kind of colors were there on any consecutive segments, he could give the number immediately without counting.

But mayor X didn't know how to give the right answer. Your friend, Mr. W was an secret officer of anti-corruption bureau, he helped mayor X on this problem and gained his trust. Do you know how Mr. Q did this?

Input

There are several test cases.

For each test case:

The first line contains two integers, N and M ,meaning that the wall is divided into N segments and there are M operations(0 < N <= 1,000,000; 0<M<=100,000)

Then M lines follow, each representing an operation. There are two kinds of operations, as described below:

1) P a b c
a, b and c are integers. This operation means that mayor X painted all segments from segment a to segment b with color c ( 0 < a<=b <= N, 0 < c <= 30).

2) Q a b
a and b are integers. This is a query operation. It means that someone asked that how many kinds of colors were there from segment a to segment b ( 0 < a<=b <= N).

Please note that the operations are given in time sequence.

The input ends with M = 0 and N = 0.

Output

For each query operation, print all kinds of color on the queried segments. For color 1, print 1, for color 2, print 2 ... etc. And this color sequence must be in ascending order.

Sample Input

5 10 P 1 2 3 P 2 3 4 Q 2 3 Q 1 3 P 3 5 4 P 1 2 7 Q 1 3 Q 3 4 P 5 5 8 Q 1 5 0 0

Sample Output

4 3 4 4 7 4 4 7 8

Source

2014 ACM/ICPC Asia Regional Guangzhou Online

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题目大意：

n段长度，m次操作。

每次操作都包涵一个op，若op为P；则随后紧跟着三个数a,b,c意味着a到b段全都渲染为c号颜色（0<c<=30）；

若op为Q，则代表着一次询问，随后的两个数字a，b代表询问a到b段所有的颜色种类（升序输出）；若某段长度不涉及P操作，

那么这段长度为2号颜色。

解题思路：

因为仅涉及30种颜色，所以仅用int型就可以表示所有的变化（二进制）；具体思路见代码注释。

# include <iostream>
# include <cstdio>
# include <cstring>
# define ls k<<1,l,mid
# define rs k<<1|1,mid+1,r

using namespace std;

const int MAXN = 1000000+10;

int n,m,ans;
int sum[MAXN];

struct node
{
	int l,r,add;
	bool is_cor;//该节点是否需要传递 
	
}tree[MAXN<<2];

void pushup(int k)//回溯，使该节点能记录其维护区间内的一切颜色 
{
    tree[k].add=(tree[k<<1].add|tree[k<<1|1].add);
    return ;
}

void pushdown(int k)//传递，将该节点区域内的颜色全部改变该节点的颜色 
{
	if(tree[k].is_cor)
	{
		int mid=k<<1;
		tree[mid].add=tree[k].add;
		tree[mid].is_cor=1;
		tree[mid|1].add=tree[k].add;
		tree[mid|1].is_cor=1;
		tree[k].is_cor=0;
	}
	return ;
}

void build(int k,int l,int r)//建树 
{
	tree[k].l=l;
	tree[k].r=r;
	tree[k].is_cor=0;
	if(l==r)
	{
		tree[k].add=2;//第2位为1，意味着2号颜色 
		return ;
	}
	int mid=(l+r)>>1;
	build(ls);
	build(rs);
	pushup(k);
	return ;
}


void updata(int k,int l,int r,int add)
{

	if(l<=tree[k].l&&tree[k].r<=r)//如果该节点全都需要改变颜色 
	{
		tree[k].is_cor=1;//需要传递 
		tree[k].add=(1<<(add-1));//第add号颜色渲染 
		return ;
	}
	pushdown(k);//传递 
	int mid = (tree[k].l+tree[k].r)>>1;
	if(r<=mid) 
		updata(k<<1,l,r,add);
	else if(l>mid)
		updata(k<<1|1,l,r,add);
	else
	{
		updata(k<<1,l,mid,add);
		updata(k<<1|1,mid+1,r,add);
	}
	pushup(k);//回溯 
	return ;
}

int query(int k,int l,int r)
{
	if(l<=tree[k].l&&tree[k].r<=r) 
	{
		return tree[k].add;
	}
	pushdown(k);
	int res=0;
	int mid=(tree[k].l+tree[k].r)>>1;
	if(r<=mid)
		res|=query(k<<1,l,r);
	else if(l>mid)
		res|=query(k<<1|1,l,r);
	else      
	{
		res|=query(k<<1,l,mid);
		res|=query(k<<1|1,mid+1,r);
	}
	return res;	
}

int main()
{
	while(~scanf("%d%d",&n,&m)&&n&&m)
	{
		build(1,1,n);
		int a,b,c;
		char ch[3];
		while(m--)
		{
			scanf("%s",ch);
			if(ch[0]=='P')
			{
				scanf("%d%d%d",&a,&b,&c);
				updata(1,a,b,c);
			}
			else
			{
				scanf("%d%d",&a,&b);
				int sz=0;
				int res=query(1,a,b);
				for(int i=0;i<30;i++)
					if(res&(1<<i))
						sum[sz++]=i+1;
				for(int i=0;i<sz-1;i++)
					cout<<sum[i]<<" ";
				cout<<sum[sz-1]<<endl;
			}
				
		}
	
	}
	return 0;
}