Count Color

Time Limit: 1000MS	Memory Limit: 65536K
Total Submissions: 13340	Accepted: 3790

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>
using namespace std;
#define MAXN 100001

struct Seg_Tree   //線段樹的結構
{
    Seg_Tree *leftptr,*rightptr;//線段樹兩個子樹的指標
    int left,right,color;//兩個節點的表示範圍和節點的顏色
};

int L,T,O,cnt;//輸入的長度L,顏色的種類T和操作的數目O
int numOfTree;//樹的當前數目
Seg_Tree tree[MAXN*4],*Root; //所有的節點和根節點

Seg_Tree* CreatNode() //建立新節點
{
    Seg_Tree *p= &tree[numOfTree++];//建立一個新的指標，指向數組裡的一個元素
    memset(p,0,sizeof(Seg_Tree));
    return p;
}

Seg_Tree* CreatTree(int s,int e)//建立樹
{
    Seg_Tree* root=CreatNode();//創立結點
    root->left=s;
    root->right=e;//為左右的範圍設立值
    if(s!=e)
    {
        int mid=(s+e)/2;//遞迴分解，節點表示的範圍越來越小
        root->leftptr=CreatTree(s,mid);//子樹表示的範圍約是父節點的1/2
        root->rightptr=CreatTree(mid+1,e);
     }
     return root;
}

bool odd(int n)//判斷這個顏色是否為單色
{
    return (n&(n-1))==0;
}

void updateTree(Seg_Tree *root,int s,int e,int color)//更新線段樹
{
    if(s<=root->left&&e>=root->right)//如果此節點表示的範圍完全被包含，則整個被重新改變顏色
    {
        root->color=color;
        return;
    }
    if(root->color==color)  return;//節點顏色不需要被改變
    if(odd(root->color))  //如果節點為單色，則在上一次改變中，子樹沒有被改變，這個重新賦值
    {
        root->leftptr->color=root->color;
        root->rightptr->color=root->color;
    }
    int mid=(root->left+root->right)/2;
    if(s<=mid) updateTree(root->leftptr,s,e,color);  //如果左子樹需要被改變
    if(e>mid) updateTree(root->rightptr,s,e,color);  //右子樹被改變
    root->color=root->leftptr->color|root->rightptr->color; //父節點的顏色等於2個子樹的或和
}

void Query(Seg_Tree* root,int s,int e,int &cnt)  //查詢
{
    if(s<=root->left&&e>=root->right) //如果節點被完全包括
    {
        cnt=cnt|root->color;
        return;
    }
    if(odd(root->color))
    {
        cnt=cnt|root->color;
        return;
    }
    int mid=(root->left+root->right)/2;
    if(s<=mid) Query(root->leftptr,s,e,cnt); //查詢左子樹
    if(e>mid) Query(root->rightptr,s,e,cnt); //查詢右子樹
}

int cal(int n) //計算顏色數量
{
    int cnt=0;
    while(n>0)
    {
        if(n%2) cnt++;
        n>>=1;
    }
    return cnt;
}

int main()
{
    numOfTree=0;
    scanf("%d%d%d",&L,&T,&O);
    Root=CreatTree(1,L);
    updateTree(Root,1,L,1);
    char cmd;
    int s,e,c;
    while (O--)
    {
        scanf(" %c",&cmd);
        if(cmd=='C')
        {
            scanf("%d%d%d",&s,&e,&c);
            if(s>e) swap(s,e);
            updateTree(Root,s,e,1<<(c-1));
        }
        else
        {
            cnt=0;
            scanf("%d%d",&s,&e);
            if(s>e) swap(s,e);
            Query(Root,s,e,cnt);
            printf("%d/n",cal(cnt));
        }
    }
    //int i;cin>>i;
    return 0;
}