poj 2777 count color 线段树

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.

```  1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 using namespace std;
5 #define lson l,m,rt<<1
6 #define rson m+1,r,rt<<1|1
7 const int MAXN=100010;
8 int cnt[MAXN<<2];
9 int lazy[MAXN<<2];
10
11 void pushup(int rt)
12 {
13     cnt[rt]=cnt[rt<<1]|cnt[rt<<1|1];
14  //   printf("cnt=%d %d %d %d\n",cnt[rt],cnt[rt<<1],cnt[rt<<1|1],rt<<1|1);
15 }
16 void pushdown(int rt)
17 {
18     if(lazy[rt])
19     {
20         cnt[rt<<1]=cnt[rt];
21         cnt[rt<<1|1]=cnt[rt];
22         lazy[rt<<1]=1;
23         lazy[rt<<1|1]=1;
24         lazy[rt]=0;
25     }
26 }
27 int query(int a,int b,int l,int r,int rt)
28 {
29    // printf("a\n");
30     //printf("query %d %d %d %d %d\n",a,b,l,r,rt);
31     if(a<=l&&b>=r) return cnt[rt];
32     pushdown(rt);
33     int t1=0,t2=0;
34     int m=(r+l)>>1;
35     if(a<=m) t1=query(a,b,lson);
36     if(b>m) t2=query(a,b,rson);
37  //   printf("query %d %d %d %d %d t1=%d t2=%d\n",a,b,l,r,rt,t1,t2);
38     return t1|t2;
39 }
40
41 void update(int a,int b,int c,int l,int r,int rt)
42 {
43    // printf("update %d %d %d %d %d %d\n",a,b,c,l,r,rt);
44     if(a<=l&&b>=r)
45     {
46         cnt[rt]=1<<(c-1);
47         lazy[rt]=1;
48       //  printf("cnt=%d l=%d r=%d c=%d\n",cnt[rt],l,r,c);
49         return;
50     }
51     pushdown(rt);
52     int m=(l+r)>>1;
53     if(a<=m) update(a,b,c,lson);
54     if(b>m) update(a,b,c,rson);
55     pushup(rt);
56 }
57 int main()
58 {
59   //  freopen("in.txt","r",stdin);
60   //  freopen("out.txt","w",stdout);
61     int len,numtype,oper,a,b,c,tmp,t,ans;
62     char type;
63     scanf("%d%d%d",&len,&numtype,&oper);
64    for(int i=0;i<MAXN<<2;i++)
65    {
66         cnt[i]=1;
67    }
68     memset(lazy,0,sizeof(lazy));
69     for(int i=1;i<=oper;i++)
70     {
71         char type[2];
72         scanf("%s",type);
73        // printf("%c\n",type[0]);
74         if(type[0]==‘C‘)
75         {
76             scanf("%d%d%d",&a,&b,&c);
77             if(a>b)
78             {
79                 t=a;a=b;b=t;
80             }
81             update(a,b,c,1,len,1);
82         }
83         else
84         {
85             scanf("%d%d",&a,&b);
86             if(a>b)
87             {
88                 t=a;a=b;b=t;
89             }
90             tmp=query(a,b,1,len,1);
91             for(ans=0;tmp>=1;tmp>>=1)
92             {
93           //  printf("tmp=%d\n",tmp);
94                 if(tmp&1) ans++;
95             }
96             printf("%d\n",ans);
97         }
98     }
99     return 0;
100 }```

POJ P2777 Count Color——线段树状态压缩

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 d

POJ 2777 Count Color （线段树区间更新加查询）

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 d

POJ 2777 Count Color

C - Count Color Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 2777 Appoint description:  System Crawler  (2015-07-22) Description Chosen Problem Solving and Program design as an optional cours

POJ 1195 2维线段树(树套树实现) 树状数组

1: #include <stdio.h> 2: #include <string.h> 3: #include <stdlib.h> 4: #include <algorithm> 5: #include <iostream> 6: using namespace std; 7:   8: #define LL(a) a<<1 9: #define RR(a) a<<1|1 10: const int MaxL = 10