POJ-2777 Count Color（线段树，区间染色问题）
POJ-2777 Count Color（线段树，区间染色问题）
Count Color Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 40510 Accepted: 12215 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:
- “C A B C” Color the board from segment A to segment B with color C.
- “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
线段树区间染色问题，用一个tag数组标记一下就好了
#include <iostream>
#include <string.h>
#include <math.h>
#include <algorithm>
#include <stdlib.h>
using namespace std;
#define MAX 100000
int cover[MAX*4+5];
int n;
int t;
int m;
char a;
int x,y,z;
int tag[31];
void PushDown(int node)
{
if(cover[node]!=-1)
{
cover[node<<1|1]=cover[node];
cover[node<<1]=cover[node];
cover[node]=-1;
}
}
void Update(int node,int begin,int end,int left,int right,int num)
{
if(left<=begin&&end<=right)
{
cover[node]=num;
return;
}
PushDown(node);
int m=(begin+end)>>1;
if(left<=m)
Update(node<<1,begin,m,left,right,num);
if(right>m)
Update(node<<1|1,m+1,end,left,right,num);
}
void Query(int node,int begin,int end,int left,int right,int &ans)
{
if(cover[node]!=-1)
{
if(!tag[cover[node]])
{
ans++;
tag[cover[node]]=1;
}
return;
}
PushDown(node);
if(begin==end)
return;
int m=(begin+end)>>1;
if(left<=m)
Query(node<<1,begin,m,left,right,ans);
if(right>m)
Query(node<<1|1,m+1,end,left,right,ans);
}
int main()
{
while(scanf("%d%d%d",&n,&t,&m)!=EOF)
{
memset(cover,0,sizeof(cover));
for(int i=1;i<=m;i++)
{
getchar();
scanf("%c",&a);
if(a=='C')
{
scanf("%d%d%d",&x,&y,&z);
Update(1,1,n,x,y,z-1);
}
else
{
memset(tag,0,sizeof(tag));
scanf("%d%d",&x,&y);
int ans=0;
Query(1,1,n,x,y,ans);
printf("%d\n",ans);
}
}
}
return 0;
}腾讯云开发者
