2777 Count Color 线段树 求区间颜色覆盖的颜色个数

Count Color

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14531		Accepted: 4137

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

 

 

 

 

 

 

 

 

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
struct Node
{
    int left,right;
    int color;
};
Node tree[1000050];
int flag[50];
void buildtree(int id,int l,int r)
{
    tree[id].left=l,tree[id].right=r;
    if(l!=r)
    {
        int mid=(l+r)/2;
        buildtree(2*id,l,mid);
        buildtree(2*id+1,mid+1,r);
    }
}
void update(int id,int l,int r,int c)
{
    if(tree[id].left==l&&tree[id].right==r)
    {
        tree[id].color=c;
        return ;
    }
    if(tree[id].color>0)
    {
        tree[2*id].color=tree[2*id+1].color=tree[id].color;//important
        tree[id].color=-1;
    }
    int mid=(tree[id].left+tree[id].right)/2;
    if(r<=mid) update(2*id,l,r,c);
    else if(l>=mid+1) update(2*id+1,l,r,c);
    else
    {
        update(2*id,l,mid,c);
        update(2*id+1,mid+1,r,c);
    }
}
void query(int id,int l,int r)
{
    if(tree[id].color>0)
    {
        flag[tree[id].color]=1;
        return ;
    }
    int mid=(tree[id].left+tree[id].right)/2;
    if(r<=mid) query(2*id,l,r);
    else if(l>=mid+1) query(2*id+1,l,r);
    else
    {
        query(2*id,l,mid);
        query(2*id+1,mid+1,r);
    }
}
int main()
{
    int n,t,o;
    while(scanf("%d%d%d",&n,&t,&o)==3)
    {
        buildtree(1,1,n);
        tree[1].color=1;
        while(o--)
        {
            char ch;cin>>ch;
            if(ch=='C')
            {
                int l,r,c;
                scanf("%d%d%d",&l,&r,&c);
                if(l>r) swap(l,r);
                update(1,l,r,c);
            }
            else
            {
                int cnt=0;memset(flag,0,sizeof(flag));
                int l,r;
                scanf("%d%d",&l,&r);
                if(l>r) swap(l,r);
                query(1,l,r);
                for(int i=1;i<=t;i++) if(flag[i]) cnt++;
                printf("%d/n",cnt);
            }
        }
    }
    return 0;
}
/*
这题有两个注意的地方，一是因为线段以segment为划分单位，所以右子树的left比左子树
的right要大1。二是给线段树结点增加一个color域，当整个线段为纯色时，color=颜色代
号，当线段内有多种颜色是，color=-1.*/