poj 2777 Count Color（線段樹+染色問題）

poj 2777 Count Color（線段樹+染色問題）

 

 

Count Color Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 40402   Accepted: 12186

 

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

題目大意：一個長度為L的區間，最多有T種顏色，並且有O種操作，接下去有o行。

一共就兩種操作：1、C a b c：表示的是將【a,b】這個區間染成顏色c。 2、P a b ：表示的是詢問【a,b】這個區間有多少種顏色。

解題思路：這個題目需要注意的是不能一直更新到最下面，就更新到符合的區間即可，否則會超時。

 

詳見代碼。

 

#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;

struct node
{
    int l,r;
    int num;
} s[400010];

int vis[35];

void InitTree(int l,int r,int k)
{
    s[k].l=l;
    s[k].r=r;
    s[k].num=1;
    int mid=(l+r)/2;
    if (l==r)
        return ;
    InitTree(l,mid,2*k);
    InitTree(mid+1,r,2*k+1);
}

void UpdataTree(int l,int r,int c,int k)
{
    if(s[k].l==l&&s[k].r==r)
    {
        s[k].num=c;
        return;
    }
    if (s[k].num==c)
        return;
    if (s[k].num!=-1)//如果所查詢的區間不是多種顏色
    {
        s[2*k].num=s[k].num;//更新區間的顏色
        s[2*k+1].num=s[k].num;
        s[k].num=-1;//-1表示該區間有多種顏色
    }
    int mid=(s[k].l+s[k].r)/2;
    if (l>mid)
        UpdataTree(l,r,c,2*k+1);
    else if (r<=mid)
        UpdataTree(l,r,c,2*k);
    else
    {
        UpdataTree(l,mid,c,2*k);
        UpdataTree(mid+1,r,c,2*k+1);
    }
}

void SearchTree(int l,int r,int k)
{
    if (s[k].num!=-1)
    {
        vis[s[k].num]=1;
        return;
    }
    int mid=(s[k].l+s[k].r)/2;
    if (r<=mid)
        SearchTree(l,r,2*k);
    else if (l>mid)
        SearchTree(l,r,2*k+1);
    else
    {
        SearchTree(l,mid,2*k);
        SearchTree(mid+1,r,2*k+1);
    }
}

int main()
{
    int l,t,o,ans;
    while (~scanf("%d%d%d",&l,&t,&o))
    {
        InitTree(1,l,1);
        while (o--)
        {
            char ch;
            int a,b,c;
            getchar();
            scanf("%c",&ch);
            if (ch=='C')
            {
                scanf("%d%d%d",&a,&b,&c);
                if (a>b)
                    swap(a,b);
                UpdataTree(a,b,c,1);
            }
            else
            {

                scanf("%d%d",&a,&b);
                if (a>b)
                    swap(a,b);
                memset(vis,0,sizeof(vis));
                SearchTree(a,b,1);
                ans=0;
                for (int i=1; i<=t; i++)
                    if (vis[i]==1)
                        ans++;
                printf ("%d\n",ans);
            }
        }
    }
    return 0;
}