POJ 2777 Count Color
Count Color
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 26788 | Accepted: 8003 |
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
Source
//不错的题目哈
//问一个区间有几种颜色
//先成段涂色 统计时遇到纯色就统计下
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#define N 100003
#define lson l,m,k<<1
#define rson m+1,r,k<<1|1
using namespace std;
int st[N<<2];
bool T[31];
void build(int l,int r,int k)
{
st[k]=1;
if(l==r)
return;
int m=(l+r)>>1;
build(lson);
build(rson);
}
void down(int &k)
{
st[k<<1]=st[k<<1|1]=st[k];
st[k]=0;
}
int cover;
void update(int &L,int &R,int l,int r,int k)
{
if(L<=l&&R>=r)
{
st[k]=cover;
return ;
}
if(st[k])
down(k);
int m=(l+r)>>1;
if(L<=m) update(L,R,lson);
if(R>m) update(L,R,rson);
}
void query(int &L,int &R,int l,int r,int k)
{ if(T[st[k]]) return ;//开始没加这句，TLE
if(L<=l&&R>=r&&st[k])
{
T[st[k]]=1;
return ;
}
if(st[k])
down(k);
int m=(l+r)>>1;
if(L<=m) query(L,R,lson);
if(R>m) query(L,R,rson);
}
int main()
{
int L,t,O;
char op;
int a,b;
int c,i;
while(scanf("%d%d%d",&L,&t,&O)!=EOF)
{
build(1,L,1);
while(O--)
{
getchar();
scanf("%c",&op);
if(op=='C')
{
scanf("%d%d%d",&a,&b,&cover);
if(a>b) swap(a,b);//这个开始也忘记加了 WA
update(a,b,1,L,1);
}
else
{
memset(T,0,sizeof(T));
scanf("%d%d",&a,&b);
if(a>b) swap(a,b);
query(a,b,1,L,1);
for(c=0,i=1;i<=t;i++)
if(T[i])
c++;
printf("%d\n",c);
}
}
}
return 0;
}
