Count Color
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 41291 | Accepted: 12481 |
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.
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 4 C 1 1 2 P 1 2 C 2 2 2 P 1 2
Sample Output
2 1
还是基础的线段树题目，和hdu 1698 非常类似。
题意：输入板子长度，颜色数量，操作次数。输入C，再输入a，b，c，表示从a到b的段刷c颜色，输入P,再输入a，b，表示询问从a到b一共有几种不同的颜色。初始全部颜色为1。
附上代码：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #define M 100005 5 using namespace std; 6 struct node 7 { 8 int l,r,s; 9 } ss[M*3]; 10 int sum[52]; 11 12 void build(int l,int r,int k) 13 { 14 ss[k].l=l; 15 ss[k].r=r; 16 ss[k].s=1; 17 if(l==r) return; 18 int mid=(l+r)/2; 19 build(l,mid,2*k); 20 build(mid+1,r,2*k+1); 21 } 22 23 void insert(int l,int r,int x,int k) 24 { 25 if(ss[k].s==x) return; 26 if(ss[k].l==l&&ss[k].r==r) 27 { 28 ss[k].s=x; 29 return; 30 } 31 if(ss[k].s!=-1) 32 { 33 ss[k*2].s=ss[k*2+1].s=ss[k].s; 34 ss[k].s=-1; 35 } 36 int mid=(ss[k].l+ss[k].r)/2; 37 if(mid>=r) insert(l,r,x,k*2); 38 else if(mid<l) insert(l,r,x,k*2+1); 39 else 40 { 41 insert(l,mid,x,k*2); 42 insert(mid+1,r,x,k*2+1); 43 } 44 } 45 void search(int l,int r,int k) 46 { 47 if(ss[k].s!=-1) 48 { 49 sum[ss[k].s]=1; 50 return; 51 } 52 else 53 { 54 int mid=(ss[k].l+ss[k].r)/2; 55 if(r<=mid) search(l,r,2*k); 56 else if(l>mid) search(l,r,2*k+1); 57 else 58 { 59 search(l,mid,2*k); 60 search(mid+1,r,2*k+1); 61 } 62 } 63 } 64 65 66 int main() 67 { 68 char ch; 69 int a,b,c,t,o,n,i; 70 while(~scanf("%d%d%d",&n,&t,&o)) 71 { 72 build(1,n,1); 73 while(o--) 74 { 75 getchar(); 76 scanf("%c",&ch); 77 if(ch=='C') 78 { 79 scanf("%d%d%d",&a,&b,&c); 80 insert(a,b,c,1); 81 } 82 else 83 { 84 scanf("%d%d",&a,&b); 85 memset(sum,0,sizeof(sum)); 86 int ans=0; 87 search(a,b,1); 88 for(i=1; i<=t; i++) 89 if(sum[i]) ans++; 90 printf("%d ",ans); 91 } 92 } 93 } 94 return 0; 95 }