poj 2777 Count Color （線段樹區間更新）

poj 2777 Count Color （線段樹區間更新）

Count Color Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 37647   Accepted: 11315

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

 

題意：

有一個board，初始都為顏色1，有兩個操作，C將區間[l,r]都塗成顏色val，P詢問區間[l,r]的顏色種數。

 

思路：

線段樹區間更新，維護一個區間的顏色種類數就行了，由於只有30種顏色，可以考慮位操作，用一個int來表示。

 

代碼：

 

#include 
#include 
#include 
#include 
#define maxn 100005
#define lson (rt<<1)
#define rson (rt<<1|1)
#define INF 0x3f3f3f3f
typedef long long ll;
using namespace std;

int n,m,tot;
char s[10];
int vis[maxn<<2],sum[maxn<<2];

void pushdown(int le,int ri,int rt)
{
    if(vis[rt]!=-1)
    {
        vis[lson]=vis[rson]=vis[rt];
        sum[lson]=sum[rson]=(1<>1;
    if(v<=mid)
    {
        update(le,mid,lson,u,v,val);
    }
    else if(u>=mid+1)
    {
        update(mid+1,ri,rson,u,v,val);
    }
    else
    {
        update(le,mid,lson,u,mid,val);
        update(mid+1,ri,rson,mid+1,v,val);
    }
    pushup(le,ri,rt);
}
int query(int le,int ri,int rt,int u,int v)
{
    if(vis[rt]!=-1) return (1<>1;
    if(v<=mid)
    {
        res=query(le,mid,lson,u,v);
    }
    else if(u>=mid+1)
    {
        res=query(mid+1,ri,rson,u,v);
    }
    else
    {
        res=query(le,mid,lson,u,mid);
        res|=query(mid+1,ri,rson,mid+1,v);
    }
    return res;
}
int main()
{
    while(~scanf("%d%d%d",&n,&tot,&m))
    {
        memset(vis,-1,sizeof(vis));
        memset(sum,0,sizeof(sum));
        update(1,n,1,1,n,0);
        while(m--)
        {
            scanf("%s",s);
            int u,v,val;
            if(s[0]=='C')
            {
                scanf("%d%d%d",&u,&v,&val);
                if(u>v) swap(u,v);
                update(1,n,1,u,v,val-1);
            }
            else
            {
                if(u>v) swap(u,v);
                scanf("%d%d",&u,&v);
                int x=query(1,n,1,u,v),ans=0;
                while(x)
                {
                    if(x&1) ans++;
                    x>>=1;
                }
                printf("%d\n",ans);
            }
        }
    }
    return 0;
}
/*
2 2 4
C 1 1 2
P 1 2
C 2 2 2
P 1 2
*/