poj 2777

Count Color

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 35280 Accepted: 10633

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

Sample Output

21

Source

POJ Monthly--2006.03.26,dodo

直接上代码了，应该看的懂吧。

AC代码：

#include<iostream>#include<stdio.h>using namespace std;struct Node{    int left,right;    int num;}a[400010];int sum;void built(int cur,int x,int y){    a[cur].left=x;    a[cur].right=y;    a[cur].num=1;    if(x==y)        return ;    int mid=(x+y)>>1;    built(cur<<1,x,mid);    built(cur<<1|1,mid+1,y);}void update(int cur,int x,int y,int val){    if(a[cur].left==x && a[cur].right==y){        a[cur].num=val;        return ;    }    if(a[cur].num){                       //当前位置全覆盖，所以要向下更新，并把本节点置0        a[cur<<1].num=a[cur<<1|1].num=a[cur].num;        a[cur].num=0;    }    int mid=(a[cur].left+a[cur].right)>>1;    if(y<=mid)        update(cur<<1,x,y,val);    else if(x>mid)        update(cur<<1|1,x,y,val);    else{        update(cur<<1,x,mid,val);        update(cur<<1|1,mid+1,y,val);    }}void query(int cur,int x,int y){    if(a[cur].num && a[cur].left<=x && y<=a[cur].right){        sum|=a[cur].num;        return ;    }    int mid=(a[cur].left+a[cur].right)>>1;    if(y<=mid)        query(cur<<1,x,y);    else if(x>mid)        query(cur<<1|1,x,y);    else{        query(cur<<1,x,mid);        query(cur<<1|1,mid+1,y);    }}int main(){    int n,t,m;    while(scanf("%d%d%d",&n,&t,&m)!=EOF){        built(1,1,n);        while(m--){            char s[2];            scanf("%s",s);            if(s[0]=='C'){                int x,y,k;                scanf("%d%d%d",&x,&y,&k);                if(x>y){                    int tmp=x;                    x=y;                    y=tmp;                }                update(1,x,y,1<<(k-1));      //因为颜色较少，所以用二进制表示颜色，方便求种类(这样种类只要取或运算得到之后看里面有多少个1就行了)            }            else{                int x,y;                scanf("%d%d",&x,&y);                if(x>y){                    int tmp=x;                    x=y;                    y=tmp;                }                sum=0;                query(1,x,y);                int ans=0;                while(sum){         //取sum里面1的个数                    if(sum&1)                        ans++;                    sum>>=1;                }                cout<<ans<<endl;            }        }    }    return 0;}