Count Color （线段树，区间更新）
Problem 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 4 C 1 1 2 P 1 2 C 2 2 2 P 1 2
Sample Output
2 1
题目大概：
在一个单行表格上，对他们进行区间涂色，最后查询区间的颜色数。
思路：
线段树，区间查询，根据别人ac代码写的。
代码：
#include
#include
#include
using namespace std;
struct poin
{
int l,r,id,v;
}a[101000<<2];
void build(int L,int R,int rt)
{
a[rt].l=L;
a[rt].r=R;
if(L==R)return;
int m=(L+R)>>1;
build(L,m,rt<<1);
build(m+1,R,rt<<1|1);
}
void update(int L,int R,int v,int rt)
{
if(L==a[rt].l&&R==a[rt].r)
{
a[rt].id=1;
a[rt].v=1<<(v-1);
return;
}
if(a[rt].id)
{
a[rt].id=0;
a[rt<<1].id=a[rt<<1|1].id=1;
a[rt<<1].v=a[rt<<1|1].v=a[rt].v;
}
int m=(a[rt].l+a[rt].r)>>1;
if(R<=m)
{
update(L,R,v,rt<<1);
}
else if(L>m)
{
update(L,R,v,rt<<1|1);
}
else
{
update(L,m,v,rt<<1);
update(m+1,R,v,rt<<1|1);
}
a[rt].v=(a[rt<<1].v|a[rt<<1|1].v);
if(a[rt<<1].id&&a[rt<<1|1].id&&a[rt<<1].v==a[rt<<1|1].v)
a[rt].id=1;
}
int quert(int L,int R,int rt)
{
if(a[rt].l==L&&a[rt].r==R)
{
return a[rt].v;
}
if(a[rt].id)
{
return a[rt].v;
}
int m=(a[rt].l+a[rt].r)>>1;
if(R<=m)
{
return quert(L,R,rt<<1);
}
else if(L>m)
{
return quert(L,R,rt<<1|1);
}
else
{
return quert(L,m,rt<<1)|quert(m,R,rt<<1|1);
}}
int main()
{
int l,k,n;
while(~scanf("%d%d%d\n",&l,&k,&n))
{build(1,l,1);
a[1].v=1;
a[1].id=1;
while(n--)
{
char q[2];
scanf("%s",q);
if(q[0]=='C')
{
int w,e,r;
scanf("%d%d%d",&w,&e,&r);
update(w,e,r,1);
}
else
{
int w,e;
scanf("%d%d",&w,&e);
int sum=quert(w,e,1);
int su=0;
for(int j=0;
j 【Count Color （线段树，区间更新）】
推荐阅读
- 2019年PB|2019年PB County Senior Open, 保龄生涯（64）
- CountDownLatch-线程并发的发令枪
- SQL优化（二）|SQL优化（二） 快速计算Distinct Count
- 24、CountIF函数和CountIFS函数
- iOS|iOS Xcode统一修改RGB的color profile
- python基础-numpy.bincount详解
- 【JUC】CountDownLatch共享节点队列
- 用Java实现线段树
- HDU 5528【2015长春现场赛 B】 Count a * b
- jQuery的弹出窗口插件colorbox