POJ2777-Count Color （线段树）

题目传送门：http://poj.org/problem?id=2777

Count Color

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 45259		Accepted: 13703

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

线段树对区间操作，简单题
给你一段L长的线让你染色，有T中颜色，共有O次命令，
命令分为
C：给x到y染成c色
P：看x到y一共有多少种颜色

#include<stdio.h>
#include<string.h>
int cc[1000];
struct A{
    int l,r,c;
}t[100000*4+4];
void bulid(int p,int l,int r)
{
    t[p].l=l;t[p].r=r;t[p].c=1;
    if(l==r)
        return ;
        int mid=(t[p].l+t[p].r)>>1;
    bulid(p*2,l,mid);
    bulid(p*2+1,mid+1,r);
}
void cha(int l,int r,int c,int p)
{
    if(t[p].l==l&&t[p].r==r)
    {
        t[p].c=c;
        return ;
    }
    if(t[p].c==c)
        return ;
    if(t[p].c!=-1)
    {
        t[p*2].c=t[p].c;
        t[p*2+1].c=t[p].c;
        t[p].c=-1;
        
    }int mid=(t[p].l+t[p].r)>>1;
        if(r<=mid)
        {
            cha(l,r,c,p*2);
        }
        else if(l>mid)
        {
            cha(l,r,c,p*2+1);
        }
        else 
        {
            cha(l,mid,c,p*2);
            cha(mid+1,r,c,p*2+1);
        }
}
void se(int l,int r,int p)
{
    if(t[p].c!=-1)
    {
        cc[t[p].c]=1;
    //    printf("!%d
",t[p].c);
        return ;
    }
    int mid=(t[p].l+t[p].r)>>1;
        if(r<=mid)
        {
            se(l,r,p*2);
        }
        else if(l>mid)
        {
            se(l,r,p*2+1);
        }
        else 
        {
            se(l,mid,p*2);
            se(mid+1,r,p*2+1);
            
        }
}
int main()
{
    int l,t,o,i,j,a,b,c;
    char aa;
    while(scanf("%d%d%d",&l,&t,&o)!=EOF)
    {
    bulid(1,1,l);
        while(o--)
        {
            getchar();
        aa=getchar();
        
        if(aa=='C')
        {
            scanf("%d%d%d",&a,&b,&c);
            if(a>b)
            {
                a=a^b;
                b=a^b;
                a=a^b;    
            }    
            cha(a,b,c,1);
        }
        else if(aa=='P')
        {
            scanf("%d%d",&a,&b);
            if(a>b)
            {
                a=a^b;
                b=a^b;
                a=a^b;    
            }
            memset(cc,0,sizeof(cc));
            se(a,b,1);
            int sun=0;
            for(j=1;j<=t;j++)
            {
                if(cc[j]==1)
                    sun++;
            }
                
                printf("%d
",sun);
                
        }
        }
        
        
    }
    return 0;    
}