Count Color
Time Limit: 1000MS Memory Limit: 65536K
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
解题心得：
- 一看就是线段树，但是还加了状压，每一段区间用一个二进制来表示其涂抹情况，假如二进制的第一位为1代表第一种颜色涂抹了，然后父节点等于两个子节点的状态或起来，询问的区间用0去线段树中或，然后数1的个数就行了。区间更新的时候lazy标记一下，但是要注意的是颜色是直接覆盖。
- 读题眼贱了一下，没看到给的left端和right端可能是交换的，然后一直Runtime Error，怀疑人生啊。
#include<stdio.h>
#include<cstring>
#include<iostream>
using namespace std;
const int maxn = 1e6+100;
struct Bitree
{
int l,r;
int co;
} bitree[maxn<<2];
int lazy[maxn<<2];
int n,a,b,c,o,t,ans;
void updata(int rt)//向上更新
{
bitree[rt].co = (bitree[rt<<1].co)|(bitree[rt<<1|1].co);//字节点或起来
}
void build_tree(int rt,int l,int r)//先初始化一棵树
{
bitree[rt].l = l;
bitree[rt].r = r;
bitree[rt].co |= 2;
if(r == l)
return ;
int mid = (l + r)>>1;
build_tree(rt<<1,l,mid);
build_tree(rt<<1|1,mid+1,r);
updata(rt);
}
void pushdown(int rt)//向下更新，注意lazy标记的转移方式就可以了
{
if(bitree[rt].l == bitree[rt].r || !lazy[rt])
return ;
bitree[rt<<1].co = bitree[rt<<1|1].co = 0;
bitree[rt<<1].co |= (1<<lazy[rt]);
bitree[rt<<1|1].co |= (1<<lazy[rt]);
lazy[rt<<1] = lazy[rt<<1|1] = lazy[rt];
lazy[rt] = 0;
}
void make_lazy(int rt,int l,int r,int L,int R)//区间更新，lazy标记
{
pushdown(rt);
if(l == L && r == R)
{
lazy[rt] = c;
bitree[rt].co = 0;
bitree[rt].co |= (1<<c);
return ;
}
int mid = (L + R) >> 1;
if(mid >= r)
make_lazy(rt<<1,l,r,L,mid);
else if(mid < l)
make_lazy(rt<<1|1,l,r,mid+1,R);
else
{
make_lazy(rt<<1,l,mid,L,mid);
make_lazy(rt<<1|1,mid+1,r,mid+1,R);
}
updata(rt);
}
void query(int rt,int l,int r,int L,int R)
{
pushdown(rt);
if(L == l && R == r)
{
ans |= bitree[rt].co;//查询的时候用ans去将线段树中的状态或出来
return ;
}
int mid = (L + R) >> 1;
if(mid >= r)
query(rt<<1,l,r,L,mid);
else if(mid < l)
query(rt<<1|1,l,r,mid+1,R);
else
{
query(rt<<1,l,mid,L,mid);
query(rt<<1|1,mid+1,r,mid+1,R);
}
updata(rt);
}
int SUM(int x)
{
int sum = 0;
while(x)
{
if(x&1)
sum++;
x >>= 1;
}
return sum;
}
int main()
{
while(cin>>n>>t>>o)
{
memset(lazy,0,sizeof(lazy));
memset(bitree,0,sizeof(bitree));
build_tree(1,1,n);
while(o--)
{
char s[10];
scanf("%s",s);
if(s[0] == 'C')
{
scanf("%d%d%d",&a,&b,&c);
if(a>b)//注意交换啊，坑死了
swap(a,b);
make_lazy(1,a,b,1,n);
}
else if(s[0] == 'P')
{
ans = 0;
scanf("%d%d",&a,&b);
if(a>b)
swap(a,b);
query(1,a,b,1,n);
ans = SUM(ans);//数最终有多少个1的个数
printf("%d
",ans);
}
}
}
return 0;
}