POJ 2777 Count Color (線段樹+狀態壓縮)

Mr_Treeeee發表於2020-04-06
Count Color
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 46327   Accepted: 14044

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

POJ Monthly--2006.03.26,dodo

題意：

L T Q，L代表數量，T代表顏色種類數量，Q代表操作次數。

C操作為 (A-B)區間內所有顏色變為C。

P操作為輸出(A-B)區間內所有顏色的種類。


POINT：

用線段樹，儲存每個區間的顏色種類，用狀態壓縮來存顏色，2的30次-int。

之後用位運算合起來就行。


#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>
using namespace std;
const int N = 100000*10;
int num[N],add[N];
void pushdown(int x)
{
    add[2*x]=add[x];
    num[2*x]=add[x];
    add[2*x+1]=add[x];
    num[2*x+1]=add[x];
    add[x]=0;
}
void change(int x,int l,int r,int ll,int rr,int c)
{
    if(add[x]) pushdown(x);
    if(ll<=l&&rr>=r)
    {
        num[x]=1<<(c-1);
        add[x]=num[x];
    }
    else
    {   int mid=(r+l)>>1;
        if(ll<=mid) change(2*x,l,mid,ll,rr,c);
        if(rr>mid) change(2*x+1,mid+1,r,ll,rr,c);
        num[x]=num[x*2+1]|num[x*2];
    }
}
int output(int x,int l,int r,int ll,int rr)
{
    int ans=0;
    if(add[x]) pushdown(x);
    if(ll<=l&&rr>=r) ans=ans|num[x];
    else
    {
        int mid=(r+l)>>1;
        if(ll<=mid) ans=ans|output(2*x,l,mid,ll,rr);
        if(rr>mid) ans=ans|output(2*x+1,mid+1,r,ll,rr);
    }
    return ans;
}
int main()
{
    int l,t,o;
    while(~scanf("%d %d %d",&l,&t,&o))
    {
        for(int i=1;i<=6*l;i++)
        {
            num[i]=1;
        }
        memset(add,0,sizeof add);
        while(o--)
        {
            char q;
            getchar();
            scanf("%c",&q);
            if(q=='C')
            {
                int a,b,c;
                scanf("%d %d %d",&a,&b,&c);
                change(1,1,l,a,b,c);
            }
            else
            {
                int a, b;
                scanf("%d %d",&a,&b);
                if(a>b) swap(a,b);
                int now=output(1,1,l,a,b);
                int ans=0;
                while(now)
                {
                    ans+=now&1;
                    now>>=1;
                }
                printf("%d\n",ans);
            }
        }
    }
}



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