POJ 2777 Count Color

C - Count Color

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d
& %I64u

Submit Status Practice POJ
2777

Appoint description: 
System Crawler  (2015-07-22)

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

有两种方法：一种是每次查询时都要统计对应区间延迟标记上颜色的种类，可以用set或简单哈希来实现。

一种是用二进制表示对应的区间涂了第几种颜色，这样每个区间除了延迟标记外，可以再开一个数组统计当前涂了哪几种颜色。这样就和一般的线段树一样了。

最后再统计一下1的数目。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define maxn 100005
#define inf 0x3f3f3f3f
typedef long long LL;
int sum[maxn<<2],col[maxn<<2],ll[maxn<<2],rr[maxn<<2];
inline void pushup(int i){
    sum[i]=sum[i<<1]|sum[i<<1|1];
}
inline void pushdown(int i){
    if(col[i]){
        col[i<<1]=col[i<<1|1]=col[i];
        sum[i<<1]=col[i];
        sum[i<<1|1]=col[i];
        col[i]=0;
    }
}
void build(int l,int r,int i){
    ll[i]=l;
    rr[i]=r;
    col[i]=1;
    if(l==r)return;
    int m=(ll[i]+rr[i])>>1,ls=i<<1,rs=ls|1;
    build(l,m,ls);
    build(m+1,r,rs);
    pushup(i);
}
void update(int l,int r,int v,int i){
    if(l<=ll[i]&&rr[i]<=r){
        col[i]=1<<(v-1);
        sum[i]=col[i];
        return ;
    }
    pushdown(i);
    int m=(ll[i]+rr[i])>>1,ls=i<<1,rs=ls|1;
    if(l<=m)update(l,r,v,ls);
    if(m<r)update(l,r,v,rs);
    pushup(i);
}
int query(int l,int r,int i){
    if(l<=ll[i]&&rr[i]<=r){
        return sum[i];
    }
    pushdown(i);
    int m=(ll[i]+rr[i])>>1,ls=i<<1,rs=ls|1;
    int ans=0;
    if(l<=m)ans=ans|query(l,r,ls);
    if(m<r)ans=ans|query(l,r,rs);
    return ans;
}
int main()
{
    int l,t,o,a,b,c;
    char q[2];
    //freopen("in.txt","r",stdin);
    while(~scanf("%d%d%d",&l,&t,&o)){
        build(1,l,1);
        for(int i=0;i<o;i++){
            scanf("%s%d%d",q,&a,&b);
            if(a>b)swap(a,b);
            if(q[0]=='C'){
                scanf("%d",&c);
                update(a,b,c,1);
            }
            else {
                int res=query(a,b,1);
                int ans=0;
                while(res){
                    if(res&1)ans++;
                    res=res>>1;
                }
                printf("%d\n",ans);
            }
        }
    }
}

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define maxn 100005
#define inf 0x3f3f3f3f
typedef long long LL;
int ll[maxn<<2],rr[maxn<<2],col[maxn<<2],vis[32];
int ans;
inline void pushdown(int i,int m){
    if(col[i]){
        col[i<<1]=col[i<<1|1]=col[i];
        col[i]=0;
    }
}
void build(int l,int r,int i){
   ll[i]=l;
   rr[i]=r;
   col[i]=1;
   if(l==r){
     return ;
   }
   int m=(ll[i]+rr[i])>>1,ls=i<<1,rs=ls|1;
   build(l,m,ls);
   build(m+1,r,rs);
}
void update(int l,int r,int v,int i){
   if(l<=ll[i]&&rr[i]<=r){
    col[i]=v;
    return ;
   }
   pushdown(i,rr[i]-ll[i]+1);
   int m=(ll[i]+rr[i])>>1,ls=i<<1,rs=ls|1;
   if(l<=m)update(l,r,v,ls);
   if(m<r) update(l,r,v,rs);
}
void query(int l,int r,int i){
   if(col[i]){
      if(!vis[col[i]]){
        ans++;
        vis[col[i]]=1;
      }
      return ;
   }
   pushdown(i,rr[i]-ll[i]+1);
   int m=(ll[i]+rr[i])>>1,ls=i<<1,rs=ls|1;
   if(l<=m)query(l,r,ls);
   if(m<r)query(l,r,rs);
}
int main()
{
    int l,t,o,a,b,c;
    char q[2];
    //freopen("in.txt","r",stdin);
    while(~scanf("%d%d%d",&l,&t,&o)){
        build(1,l,1);
        for(int i=0;i<o;i++){
            scanf("%s%d%d",q,&a,&b);
            if(a>b)swap(a,b);
            if(q[0]=='C'){
                scanf("%d",&c);
                update(a,b,c,1);
            }
            else {
                ans=0;
                memset(vis,0,sizeof vis);
                query(a,b,1);
                printf("%d\n",ans);
            }
        }
    }
}

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