POJ2777——Count Color(线段树)

Count Color

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 33637		Accepted: 10153

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

第一道线段树题目，未完待续……有空再写

#include <algorithm>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#define INF 0x7fffffff
#define MAX 100000
using namespace std;
struct SegTree
{
    int left,right,color;
}tree[MAX*4];
bool vis[50];
int sum=0;
void Build(int left,int right,int k)   // [l,mid],[mid+1,r] 只含整数点
{
    tree[k].left=left;
    tree[k].right=right;
    tree[k].color=1;
    if(left==right) return;

    int mid=(left+right)/2;
    Build(left,mid,k*2);
    Build(mid+1,right,k*2+1);
}
void Insert(int st,int ed,int color,int k)   //插入线段
{
    if(tree[k].left==st&&tree[k].right==ed)  // 如果区间符合，涂色，返回
    {
        tree[k].color=color;
        return ;
    }
    if(tree[k].color)  // 标记下移
    {
        tree[k*2].color=tree[k].color;
        tree[k*2+1].color=tree[k].color;
        tree[k].color=0;
    }
    int mid = (tree[k].left+tree[k].right)/2 ;  //找到中间值
    //判断插入线段位置
    if(mid>=ed)                       //插入左子树
        Insert(st,ed,color,k*2);
    else if(mid<st)                   //插入右子树
        Insert(st,ed,color,k*2+1);
    else                              //分段，插入左右子树
    {
        Insert(st,mid,color,k*2);
        Insert(mid+1,ed,color,k*2+1);
    }
}
void Query(int st,int ed,int k)   //询问操作
{
    if(tree[k].color)
    {
        if(!vis[tree[k].color])
        {
            vis[tree[k].color]=1;
            sum++;
        }
        return ;
    }
    int mid=(tree[k].left+tree[k].right)/2;
    if(mid>=ed)
        Query(st,ed,k*2);
    else if(mid<st)
        Query(st,ed,k*2+1);
    else
    {
        Query(st,mid,k*2);
        Query(mid+1,ed,k*2+1);
    }
}
int main()
{
    int n,m,c;
    char q[3];
    int l,r,cl;
    cin>>n>>m>>c;
    Build(1,n,1);
    while(c--)
    {
        scanf("%s",q);
        if(q[0]=='C')
        {
            scanf("%d%d%d",&l,&r,&cl);
            if(l>r) swap(l,r);
            Insert(l,r,cl,1);
        }
        else if(q[0]=='P')
        {
            sum=0;
            scanf("%d%d",&l,&r);
            if(l>r) swap(l,r);
            memset(vis,0,sizeof(vis));
            Query(l,r,1);
            cout<<sum<<endl;
        }
    }
    return 0;
}