POJ 2777-Count Color(线段树_区间染色)

Count Color

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ
2777

Appoint description: 
System Crawler  (2015-04-10)

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

利用二进制表示颜色。T只有30种，所以妥妥的。

#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <sstream>
#include <algorithm>
#include <set>
#include <queue>
#include <stack>
#include <map>
using namespace std;
typedef long long LL;
const int inf=0x3f3f3f3f;
const double pi= acos(-1.0);
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
const int MAXN=100010;
int sum[MAXN<<3];
int lazy[MAXN<<3];
void PushUp(int rt)
{
    sum[rt]=sum[rt<<1]|sum[rt<<1|1];
}

void PushDown(int rt)
{
    if(lazy[rt]) {
        lazy[rt<<1]=lazy[rt];
        lazy[rt<<1|1]=lazy[rt];
        sum[rt<<1]=lazy[rt];
        sum[rt<<1|1]=lazy[rt];
        lazy[rt]=0;
    }
}

void Update(int ll, int rr, int x, int l, int r, int rt)
{
    if(ll<=l&&rr>=r) {
        sum[rt]=x;
        lazy[rt]=x;
        return ;
    }
    PushDown(rt);
    int mid=l+r>>1;
    if(ll<=mid)
        Update(ll,rr,x,lson);
    if(rr>mid)
        Update(ll,rr,x,rson);
    PushUp(rt);
}
int Query(int ll, int rr, int l, int r, int rt)
{
    if(ll<=l&&rr>=r) {
        return sum[rt];
    }
    PushDown(rt);
    int ans=0;
    int mid=l+r>>1;
    if(ll<=mid)
        ans=ans|Query(ll,rr,lson);
    if(rr>mid)
        ans=ans|Query(ll,rr,rson);
    return ans;
}
int get(int x)
{
    int ans=0;
    int y;
    while(x) {
        y=x%2;
        if(y)
            ans++;
        x/=2;
    }
    return ans;
}
int main()
{
    int n, i,t, q;
    int a,b,c;
    char ch;
    scanf("%d %d %d",&n,&t,&q);
    memset(lazy,0,sizeof(lazy));
    for(i=1; i<=4*n; i++) {
        sum[i]=1;
    }
    while(q--) {
        getchar();
        scanf("%c",&ch);
        if(ch=='C') {
            scanf("%d %d %d",&a,&b,&c);
            if(a>b)
                swap(a,b);
            Update(a,b,1<<(c-1),1,n,1);
        } else {
            scanf("%d %d",&a,&b);
            if(a>b)
                swap(a,b);
            int ans=get(Query(a,b,1,n,1));
            printf("%d\n",ans);
        }
    }
    return 0;
}