poj 2777 Count Color （线段树区间更新）

最新推荐文章于 2022-12-02 10:34:03 发布

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最新推荐文章于 2022-12-02 10:34:03 发布

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本文介绍了一道关于线段树的经典问题，通过区间更新和查询不同颜色的数量，展示了如何利用位操作优化数据结构，实现高效的颜色种类计数。

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Count Color

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 37647		Accepted: 11315

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

Sample Output

2
1

Source

POJ Monthly--2006.03.26,dodo

题意：

有一个board，初始都为颜色1，有两个操作，C将区间[l,r]都涂成颜色val，P询问区间[l,r]的颜色种数。

思路：

线段树区间更新，维护一个区间的颜色种类数就行了，由于只有30种颜色，可以考虑位操作，用一个int来表示。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define maxn 100005
#define lson (rt<<1)
#define rson (rt<<1|1)
#define INF 0x3f3f3f3f
typedef long long ll;
using namespace std;

int n,m,tot;
char s[10];
int vis[maxn<<2],sum[maxn<<2];

void pushdown(int le,int ri,int rt)
{
    if(vis[rt]!=-1)
    {
        vis[lson]=vis[rson]=vis[rt];
        sum[lson]=sum[rson]=(1<<vis[rt]);
        vis[rt]=-1;
    }
}
void pushup(int le,int ri,int rt)
{
    sum[rt]=sum[lson]|sum[rson];
}
void update(int le,int ri,int rt,int u,int v,int val)
{
    if(le==u&&ri==v)
    {
        vis[rt]=val;
        sum[rt]=(1<<val);
        return ;
    }
    pushdown(le,ri,rt);
    int mid=(le+ri)>>1;
    if(v<=mid)
    {
        update(le,mid,lson,u,v,val);
    }
    else if(u>=mid+1)
    {
        update(mid+1,ri,rson,u,v,val);
    }
    else
    {
        update(le,mid,lson,u,mid,val);
        update(mid+1,ri,rson,mid+1,v,val);
    }
    pushup(le,ri,rt);
}
int query(int le,int ri,int rt,int u,int v)
{
    if(vis[rt]!=-1) return (1<<vis[rt]);
    else if(le==u&&ri==v)
    {
        return sum[rt];
    }
    pushdown(le,ri,rt);
    int res=0,mid=(le+ri)>>1;
    if(v<=mid)
    {
        res=query(le,mid,lson,u,v);
    }
    else if(u>=mid+1)
    {
        res=query(mid+1,ri,rson,u,v);
    }
    else
    {
        res=query(le,mid,lson,u,mid);
        res|=query(mid+1,ri,rson,mid+1,v);
    }
    return res;
}
int main()
{
    while(~scanf("%d%d%d",&n,&tot,&m))
    {
        memset(vis,-1,sizeof(vis));
        memset(sum,0,sizeof(sum));
        update(1,n,1,1,n,0);
        while(m--)
        {
            scanf("%s",s);
            int u,v,val;
            if(s[0]=='C')
            {
                scanf("%d%d%d",&u,&v,&val);
                if(u>v) swap(u,v);
                update(1,n,1,u,v,val-1);
            }
            else
            {
                if(u>v) swap(u,v);
                scanf("%d%d",&u,&v);
                int x=query(1,n,1,u,v),ans=0;
                while(x)
                {
                    if(x&1) ans++;
                    x>>=1;
                }
                printf("%d\n",ans);
            }
        }
    }
    return 0;
}
/*
2 2 4
C 1 1 2
P 1 2
C 2 2 2
P 1 2
*/