POJ 2777 Count Color 【线段树 区间更新 按位或运算】

Count Color

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 40183 Accepted: 12127

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.

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

恩，题目大意就是说，一块木板，初始颜色为1号，然后给不同的区间染色，问知道区间内的颜色个数。刚开始做，各种不对啊，看人家题解，里面的按位或运算完全不懂啊，不懂啊。。。。。

#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>#define maxn 100000using namespace std;int l,t,q,sum=0;struct lnode{    int l,r,c;    bool f;};lnode node[maxn<<2];void pushup(int o){    node[o].c=node[o<<1].c|node[o<<1|1].c;}void build(int o,int l,int r){    node[o].l=l;    node[o].r=r;    node[o].c=1;    node[o].f=true;    if(l==r)        return ;    int mid=(l+r)>>1;    build(o<<1,l,mid);    build(o<<1|1,mid+1,r);}void pushdown(int o){    if(node[o].f)    {        node[o<<1].c=node[o<<1|1].c=node[o].c;        node[o<<1].f=node[o<<1|1].f=true;        node[o].f=false;    }}void update(int o,int l,int r,int x){    if(node[o].l==l&&node[o].r==r)    {        node[o].c=x;        node[o].f=true;        return ;    }    if(node[o].c==x)        return ;    pushdown(o);    int mid=(node[o].l+node[o].r)>>1;    if(r<=mid)        update(o<<1,l,r,x);    else if(l>mid)        update(o<<1|1,l,r,x);    else    {        update(o<<1,l,mid,x);        update(o<<1|1,mid+1,r,x);    }    pushup(o);}void query(int o,int l,int r){    if(node[o].l>=l&&node[o].r<=r)    {        sum|=node[o].c;        return ;    }    if(node[o].f)    {        sum|=node[o].c;        return ;    }    int mid=(node[o].l+node[o].r)>>1;    if(r<=mid)        query(o<<1,l,r);    else if(l>mid)        query(o<<1|1,l,r);    else    {        query(o<<1,l,mid);        query(o<<1|1,mid+1,r);    }}int get(int sum){    int ans=0;    while(sum)    {        if(sum&1)            ans++;        sum>>=1;    }    return ans;}int main(){    int a,b,x;    char c;    while(~scanf("%d%d%d",&l,&t,&q))    {        build(1,1,l);        while(q--)        {            getchar();            scanf("%c",&c);            sum=0;            if(c=='C')            {                scanf("%d%d%d",&a,&b,&x);                if(a>b)                {                    int tem=a;                    a=b;                    b=tem;                }                update(1,a,b,1<<(x-1));            }            else            {                scanf("%d%d",&a,&b);                if(a>b)                {                    int tem=a;                    a=b;                    b=tem;                }                query(1,a,b);                printf("%d\n",get(sum));            }        }    }    return 0;}