POJ 2777-Count Color（線段樹-區間染色查詢）

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. 

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.

Ouput results of the output operation in order, each line contains a number.

2 2 4
C 1 1 2
P 1 2
C 2 2 2
P 1 2

2
1

POJ Monthly--2006.03.26,dodo

#include<iostream>
#include<cstdio>
#include<iomanip>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<map>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
#define INF 0xfffffff
#define MAXN 275000
bool vis[50];//1表示該區間有該顏色
int ans[MAXN*4];
struct Node //樹
{
    int l,r;//左右節點
    int col;//區間內的顏色
};
Node tree[MAXN*5];

void BuildTree(int root, int l, int r)//建樹
{
    tree[root].l=l;
    tree[root].r=r;
    tree[root].col=1;//初始狀態都是同一種顏色
    if(l!=r)
    {
        BuildTree(2*root,l,(l+r)/2);
        BuildTree(2*root+1,(l+r)/2+1,r);
    }
}
void Insert(int i,int l,int r,int c)//插入
{
    if(tree[i].col==c)//同色
        return ;
    if(tree[i].l==l&&tree[i].r==r)
    {
        tree[i].col=c;//染色
        return ;
    }
    if(tree[i].col!=-1)//同色
    {
        tree[2*i].col=tree[2*i+1].col=tree[i].col;
        tree[i].col=-1;
    }
    int mid=(tree[i].l+tree[i].r)/2;
    if(r<=mid)
        Insert(2*i,l,r,c);
    else if(l>mid)
        Insert(2*i+1,l,r,c);
    else
    {
        Insert(2*i,l,mid,c);
        Insert(2*i+1,mid+1,r,c);
    }
}
void Query(int i,int l,int r)
{
    if(tree[i].col!=-1)//同色
    {
        vis[tree[i].col]=true;
        return;
    }
    int mid=(tree[i].l+tree[i].r)/2;
    if(r<=mid)
        Query(2*i,l,r);
    else if(l>mid)
        Query(2*i+1,l,r);
    else
    {
        Query(2*i,l,mid);
        Query(2*i+1,mid+1,r);
    }
}

int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("G:/cbx/read.txt","r",stdin);
    //freopen("G:/cbx/out.txt","w",stdout);
#endif
    int l,t,n;
    scanf("%d%d%d\n",&l,&t,&n);
    //while(~scanf("%d%d%d\n",&l,&t,&n))
    {
        BuildTree(1,1,n);//建樹
        while(n--)
        {
            char ch;
            cin>>ch;
            if(ch=='C')
            {
                int a,b,c;
                scanf("%d%d%d",&a,&b,&c);
                if(a>b) swap(a,b);//如果a>b
                Insert(1,a,b,c);
            }
            else if(ch=='P')
            {
                memset(vis,false,sizeof(vis));
                int a,b,ans=0;
                scanf("%d%d",&a,&b);
                if(a>b) swap(a,b);
                Query(1,a,b);
                for(int i=1; i<=t; ++i)
                    if(vis[i]) ++ans;
                printf("%d\n",ans);
            }
        }
    }
    return 0;
}