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poj 2777 Count Color

2024-07-08 23:20:23   228人阅读

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

一道简单的线段树区间更新，区间查询的题目，当然是要用mark标记，网上好多人说这是laxy标记，其实这不是lazy，而是更好的方法，主要是30种颜色怎么表示，当然好表示了，2的30次放也是整形数据嘛，用集合的交并就可以了
#include<map>#include<set>#include<queue>#include<cmath>#include<vector>#include<cstdio>#include<string>#include<cstring>#include<cstdlib>#include<iostream>#include<algorithm>#define  inf 0x0f0f0f0fusing namespace std;const int maxn=100000+10;struct node{     int left,right,sum,mark;}tree[maxn*8];int get_sum(int s){     int ans=0;     while(s)     {          if (s%2) ans++;          s=s/2;     }     return ans;}void build(int c,int x,int y){     tree[c].left=x; tree[c].right=y; tree[c].sum=1; tree[c].mark=0;     if (x==y) return;     int mid=x+(y-x)/2;     build(c*2,x,mid);     build(c*2+1,mid+1,y);}void update(int c,int x,int y,int v){     if (tree[c].left==x && tree[c].right==y)     {          tree[c].sum=1<<v;          tree[c].mark=v+1;          return;     }     if (tree[c].mark)     {          tree[c*2].sum=1<<(tree[c].mark-1);          tree[c*2].mark=tree[c].mark;          tree[c*2+1].sum=1<<(tree[c].mark-1);          tree[c*2+1].mark=tree[c].mark;          tree[c].mark=0;     }     int mid=tree[c].left+(tree[c].right-tree[c].left)/2;     if (y<=mid) update(c*2,x,y,v);     else if (x>mid) update(c*2+1,x,y,v);     else     {          update(c*2,x,mid,v);          update(c*2+1,mid+1,y,v);     }     tree[c].sum=tree[c*2].sum | tree[c*2+1].sum;}int reseach(int c,int x,int y){     if (tree[c].left==x && tree[c].right==y)     return tree[c].sum;     if (tree[c].mark)     {          tree[c*2].sum=1<<(tree[c].mark-1);          tree[c*2].mark=tree[c].mark;          tree[c*2+1].sum=1<<(tree[c].mark-1);          tree[c*2+1].mark=tree[c].mark;          tree[c].mark=0;     }     int mid=tree[c].left+(tree[c].right-tree[c].left)/2;     if (y<=mid) return reseach(c*2,x,y);     else if (x>mid) return reseach(c*2+1,x,y);     else     {          return (reseach(c*2,x,mid) | reseach(c*2+1,mid+1,y));     }}int main(){     int n,t,m,x,y,v;     char str[3];     while(scanf("%d%d%d",&n,&t,&m)!=EOF)     {          build(1,1,n);          while(m--)          {               scanf("%s",str);               if (str[0]==‘C‘)               {                    scanf("%d%d%d",&x,&y,&v);                    if (x>=y) swap(x,y);                    update(1,x,y,v-1);               }               else               {                    scanf("%d%d",&x,&y);                    if (x>=y) swap(x,y);                    int temp=reseach(1,x,y);                    int ans=get_sum(temp);                    printf("%d\n",ans);               }          }     }     return 0;}

作者 chensunrise


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