POJ 2777 Count Color 线段树+小技巧

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最新推荐文章于 2022-01-29 12:06:18 发布

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分类专栏： ACM-数据结构

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本文介绍了一种使用线段树数据结构解决特定涂色问题的方法。问题要求在一长度为L的色板上进行涂色操作并统计不同颜色的数量。通过构建线段树并实现更新和查询操作，高效地解决了该问题。

Count Color

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 44939		Accepted: 13614

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：

简单英文，很容易直到意思，就是涂颜色嘛。

解题思路：

一看就是裸的线段树，T给的30很明显就是要你用数位来计数，想到这就很好做了。直接板子一套。

#include<iostream>
#include<cstdio>
#include<map>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
const int maxn = 100005 ;
int col[maxn<<2] ;
int sum[maxn<<2] ;
int L,T,O ;
int n ;
void PushUp(int rt){
    sum[rt] = sum[rt<<1]|sum[rt<<1|1] ;
    return ;
}
void PushDown(int rt){
    if(col[rt]){
        col[rt<<1] = col[rt] ;
        col[rt<<1|1] = col[rt] ;
        sum[rt<<1] = col[rt] ;
        sum[rt<<1|1] = col[rt] ;
        col[rt]=0;
    }
    return ;
}
void build(int l,int r,int rt){
    col[rt]=0;
    sum[rt]=0;
    if(l==r)return ;
    int m = (l+r)>>1 ;
    build(lson) ;
    build(rson) ;
    PushUp(rt);
}
void update(int L,int R,int c,int l,int r,int rt){
    if(L<=l&&r<=R){
        col[rt]=c ;
        sum[rt] = c ;
        return ;
    }
    PushDown(rt);
    int m = (l+r)>>1 ;
    if(L<=m)update(L,R,c,lson) ;
    if(m<R)update(L,R,c,rson) ;
    PushUp(rt) ;
}
int query(int L,int R,int l,int r,int rt){
    if(L<=l&&r<=R){
        return sum[rt] ;
    }
    PushDown(rt) ;
    int m = (l+r)>>1 ;
    int ret =0;
    if(L<=m)ret |= query(L,R,lson) ;
    if(m<R) ret |= query(L,R,rson) ;
    PushUp(rt) ;
    return ret ;
}
int main(){
    map<int,int>mapp;
    while(~scanf("%d%d%d",&L,&T,&O)){
        mapp.clear() ;
        n=L ;
        build(1,n,1) ;
        int cnt=2 ;
        update(1,n,1,1,n,1) ;
        mapp[1]=1 ;
        for(int i=0;i<O;i++){
            char op[3];
            scanf("%s",op);
            if(op[0]=='C'){
                int a,b,c ;
                scanf("%d%d%d",&a,&b,&c);
                if(a>b)swap(a,b) ;
                if(mapp.count(c)==0){
                    mapp[c]=cnt ;
                    cnt<<=1 ;
                    //printf("cnt = %d\n",cnt);
                }
                update(a,b,mapp[c],1,n,1) ;
//                for(int j=1;j<=4;j++){
//                    printf("%d ",sum[j]);
//                }printf("\n");
            }else{
                int a,b ;
                scanf("%d%d",&a,&b);
                if(a>b)swap(a,b) ;
                int x = query(a,b,1,n,1) ;
                //printf("x = %d\n",x);
                int ans = 0 ;
                while(x){
                    if(x%2==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

*/