Count Color（线段树+染色问题）

Count Color

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 42   Accepted Submission(s) : 13

Problem 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

PKU

题目大意：一个长度为L的区间，最多有T种颜色，并且有O种操作，然后接下去有o行，表示的o行操作，一共就两种操作：第一种是  C a b c：表示的是将[a,b]这个区间染成颜色c。 第二种是 P a b ：表示的是询问[a,b]这个区间有多少种颜色，然后输出。

解题思路：典型的线段树题目，自己还是有点手生，不能用cin cout，否则会超时。这个题目典型的就是线段树加染色问题，这是很经典的问题。这个题目需要注意的是不能一直更新到最下面，就更新到符合的区间即可，否则会超时。

代码：

#include <iostream>#include <cstring>#include <cstdio>#include <stdio.h>#include <algorithm>using namespace std;int vis[38];struct node{int l,r;int num;}s[400005];void buildtree(int l,int r,int k)  //建立一个二叉树{s[k].l=l;s[k].r=r;s[k].num=1;int mid=(l+r)/2;if(l==r)return;buildtree(l,mid,2*k);buildtree(mid+1,r,2*k+1);}void update(int l,int r,int c,int k)  //对区间进行更新操作，注意不是更新到最下面{if(s[k].l==l&&s[k].r==r){s[k].num=c;return;}if(s[k].num==c)return;if(s[k].num!=-1){                       //如果所查询的区间不是多种颜色        s[2*k].num=s[k].num;//更新区间的颜色s[2*k+1].num=s[k].num;s[k].num=-1;//-1表示有多种颜色}int mid=(s[k].l+s[k].r)/2;if(l>mid)update(l,r,c,2*k+1);else if(r<=mid)update(l,r,c,2*k);else {update(l,mid,c,2*k);update(mid+1,r,c,2*k+1);}}void search(int l,int r,int k){if(s[k].num!=-1){vis[s[k].num]=1;return;} int mid=(s[k].l+s[k].r)/2;      if (r<=mid)          search(l,r,2*k);      else if (l>mid)          search(l,r,2*k+1);      else      {          search(l,mid,2*k);          search(mid+1,r,2*k+1);      }  }int main(){//ios::sync_with_stdio(false);int l,t,o,ans;      while (~scanf("%d%d%d",&l,&t,&o))      {         buildtree(1,l,1);          while (o--)          {              char ch;              int a,b,c;              getchar();scanf("%c",&ch);            if (ch=='C')              {                  scanf("%d%d%d",&a,&b,&c);                if (a>b)  swap(a,b);                  update(a,b,c,1);              }              else              {                    scanf("%d%d",&a,&b);                if (a>b)  swap(a,b);                  memset(vis,0,sizeof(vis));                  search(a,b,1);                  ans=0;                  for (int i=1; i<=t; i++)                      if (vis[i]==1)  ans++;               printf ("%d\n",ans);              }          }      }      return 0;  }

线段树做起来不顺手，还得多做题。