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poj2777Count Color（線段樹）

編輯：關於C語言

poj2777Count Color（線段樹）

Count Color Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 40404 Accepted: 12188

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

Sample Output

2
1

Source

POJ Monthly--2006.03.26,dodo
題意：一條很長（L）的畫板，有T種顏色，O個操作；每次操作將一個區間刷成一種顏色，或者查詢一個區間內所含的顏色數。分析：這題和hud1698很像，就查詢有所不同，具體看代碼和HDU1698。

#include 
#include 
#include 
#include 
#include 
#include
#include 
#include 
#include 
#include 
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
#define ll long long
#define CL(a,b) memset(a,b,sizeof(a))
#define MAXN 100010

struct node
{
    int l,r,s;
} t[MAXN<<2];
int L,O,T;
int sum[33];//因為顏色數不多，可以用一個數組來裝；1表示該區間有該顏色，0反之

void build(int l, int r, int i)
{
    t[i].l = l;
    t[i].r = r;
    t[i].s = 1;
    if(l == r) return ;
    int mid = (l+r)>>1;
    build(l, mid, i<<1);
    build(mid+1, r, i<<1|1);
}

void update(int l, int r, int m, int i)
{
    if(t[i].s == m) return ;
    if(t[i].l == l && t[i].r == r)
    {
        t[i].s = m;
        return ;
    }
    if(t[i].s != -1)
    {
        t[i<<1].s = t[i<<1|1].s = t[i].s;
        t[i].s = -1;
    }
    int mid = (t[i].l+t[i].r)>>1;
    if(l > mid) update(l, r, m, i<<1|1);
    else if(r <= mid) update(l, r, m, i<<1);
    else
    {
        update(l, mid, m, i<<1);
        update(mid+1, r, m, i<<1|1);
    }
}

void query(int l, int r, int i)
{
    if(t[i].s != -1)//如果純色把該顏色標記
    {
        sum[t[i].s] = 1;
        return ;
    }
    else//否則繼續查詢子節點
    {
        int mid = (t[i].l+t[i].r)>>1;
        if(l > mid) query(l, r, i<<1|1);
        else if(r <= mid) query(l, r, i<<1);
        else
        {
            query(l, mid, i<<1);
            query(mid+1, r, i<<1|1);
        }
    }
}

int main()
{
    char ch;
    int a,b,c;
    while(scanf("%d%d%d",&L,&T,&O)==3)
    {
        build(1, L, 1);
        while(O--)
        {
            getchar();
            scanf("%c",&ch);
            if(ch == 'C')
            {
                scanf("%d%d%d",&a,&b,&c);
                update(a, b, c, 1);
            }
            int ans = 0;
            if(ch == 'P')
            {
                CL(sum, 0);//每次查詢之前清零
                scanf("%d%d",&a,&b);
                query(a, b, 1);
                for(int i=1; i<=T; i++)//統計出現過的顏色數
                    if(sum[i]) ans++;
                printf("%d\n",ans);
            }
        }
    }
    return 0;
}