【POJ 2777】 Count Color（线段树区间更新与查询）-白红宇

【POJ 2777】 Count Color（线段树区间更新与查询）

阅读量：4599 次

发布时间：2019-06-09

本文共 3409 字，大约阅读时间需要 11 分钟。

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 40949		Accepted: 12366

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

Source

,dodo

感觉有些神经衰弱了。。。写发水水的线段树放松放松。。。

一直在搞图论是该换换弦磨得快锈了太折磨人了……（弱也不知道说了些啥

一块木板长L(1~L) 有T种颜色的油漆标号1~T 默认木板初始是1号颜色

进行O次操作操作有两种

C a b c 表示木板a~b段涂c种油漆（若之前涂过其它颜色则覆盖掉）

P a b 表示询问木板a~b段如今涂了几种油漆

两个数组一个存树一个存涂了哪几种油漆

存树的表示a~b涂的某种颜色然后搞搞就出来了……好乏

代码例如以下：

#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            
             #include 
             
              #include 
              
               #include 
               #include 
                
                 #define LL long long#define Pr pair
                 
                  #define fread() freopen("in.in","r",stdin)#define fwrite() freopen("out.out","w",stdout)using namespace std;const int INF = 0x3f3f3f3f;const int msz = 10000;const int mod = 1e9+7;const double eps = 1e-8;int tr[400400];bool col[33];int L,T,O;void Color(int root,int l,int r,int a,int b,int c){ if(a == l && b == r) { tr[root] = c; return; } int mid = (l+r)>>1; if(tr[root]) tr[root<<1] = tr[root<<1|1] = tr[root]; tr[root] = 0; if(mid >= b) Color(root<<1,l,mid,a,b,c); else if(mid+1 <= a) Color(root<<1|1,mid+1,r,a,b,c); else { Color(root<<1,l,mid,a,mid,c); Color(root<<1|1,mid+1,r,mid+1,b,c); }}void Search(int root,int l,int r,int a,int b){ //printf("root:%d l:%d r:%d co:%d\n",root,l,r,tr[root]); if(tr[root]) { col[tr[root]] = 1; return; } int mid = (l+r)>>1; if(mid >= b) Search(root<<1,l,mid,a,b); else if(mid+1 <= a) Search(root<<1|1,mid+1,r,a,b); else { Search(root<<1,l,mid,a,mid); Search(root<<1|1,mid+1,r,mid+1,b); }}int main(){ //fread(); //fwrite(); char opt[3]; int a,b,c; while(~scanf("%d%d%d",&L,&T,&O)) { memset(tr,0,sizeof(tr)); tr[1] = 1; while(O--) { scanf("%s",opt); scanf("%d%d",&a,&b); if(a > b) swap(a,b); if(opt[0] == 'C') { scanf("%d",&c); Color(1,1,L,a,b,c); } else { memset(col,0,sizeof(col)); Search(1,1,L,a,b); int ans = 0; for(int i = 1; i <= T; ++i) ans += col[i]; printf("%d\n",ans); } } } return 0;}

转载于:https://www.cnblogs.com/lxjshuju/p/7105277.html

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