Problem Description
Do you like painting? Little D doesn't like painting, especially messy color paintings. Now Little B is painting. To prevent him from drawing messy painting, Little D asks you to write a program to maintain following operations. The specific format of these operations is as follows.
0 : clear all the points.
1 x y c : add a point which color is c at point (x,y).
2 x y1 y2 : count how many different colors in the square (1,y1) and (x,y2). That is to say, if there is a point (a,b) colored c, that 1≤a≤x and y1≤b≤y2, then the color c should be counted.
3 : exit.
Input
The input contains many lines.
Each line contains a operation. It may be '0', '1 x y c' ( 1≤x,y≤106,0≤c≤50 ), '2 x y1 y2' (1≤x,y1,y2≤106 ) or '3'.
x,y,c,y1,y2 are all integers.
Assume the last operation is 3 and it appears only once.
There are at most 150000 continuous operations of operation 1 and operation 2.
There are at most 10 operation 0.
Output
For each operation 2, output an integer means the answer .
Sample Input
0
1 1000000 1000000 50
1 1000000 999999 0
1 1000000 999999 0
1 1000000 1000000 49
2 1000000 1000000 1000000
2 1000000 1 1000000
0
1 1 1 1
2 1 1 2
1 1 2 2
2 1 1 2
1 2 2 2
2 1 1 2
1 2 1 3
2 2 1 2
2 10 1 2
2 10 2 2
0
1 1 1 1
2 1 1 1
1 1 2 1
2 1 1 2
1 2 2 1
2 1 1 2
1 2 1 1
2 2 1 2
2 10 1 2
2 10 2 2
3
Sample Output
2
3
1
2
2
3
3
1
1
1
1
1
1
1
Description(CHN)
(1~x~y~c) ：在 ((x,y)) 点涂上c这种颜色
(2~x~y_1~y_2) ：从矩形左下角 ((1,y_1)) 到矩形右上角 ((x,y_2)) 范围内有多少种颜色
(0) ：清空
(3) ：表示退出
Solution
考虑对每一种颜色建一棵线段树
询问时枚举每一种颜色，看是否存在矩阵中
因为询问一定是从 (x=1) 开始，所以对于一种颜色，线段树位置 (i) 表示矩形 (y=i) 的那一列中当前颜色出现的最小 (x) 是什么，再与询问的条件比较即可
空间开不下，用动态开点
#include<bits/stdc++.h>
#define ui unsigned int
#define ll long long
#define db double
#define ld long double
#define ull unsigned long long
const int MAXN=1000000+10,inf=0x3f3f3f3f;
int opt,N=MAXN,C=50,na;
template<typename T> inline void read(T &x)
{
T data=0,w=1;
char ch=0;
while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
if(ch=='-')w=-1,ch=getchar();
while(ch>='0'&&ch<='9')data=((T)data<<3)+((T)data<<1)+(ch^'0'),ch=getchar();
x=data*w;
}
template<typename T> inline void write(T x,char ch=' ')
{
if(x<0)putchar('-'),x=-x;
if(x>9)write(x/10);
putchar(x%10+'0');
if(ch!=' ')putchar(ch);
}
template<typename T> inline void chkmin(T &x,T y){x=(y<x?y:x);}
template<typename T> inline void chkmax(T &x,T y){x=(y>x?y:x);}
template<typename T> inline T min(T x,T y){return (x<y?x:y);}
template<typename T> inline T max(T x,T y){return (x>y?x:y);}
#define Mid ((l+r)>>1)
#define lson l,Mid
#define rson Mid+1,r
struct Segment_Tree{
int Mn[MAXN<<2],lc[MAXN<<2],rc[MAXN<<2],cnt,root[50+10];
inline void init()
{
for(register int i=0;i<=cnt;++i)lc[i]=rc[i]=0;
memset(root,0,sizeof(root));
cnt=0;
}
inline void Update(int &rt,int l,int r,int ps,int k)
{
if(!rt)Mn[rt=++cnt]=k;
else chkmin(Mn[rt],k);
if(l==r)return ;
else
{
if(ps<=Mid)Update(lc[rt],lson,ps,k);
else Update(rc[rt],rson,ps,k);
}
}
inline void Query(int rt,int l,int r,int L,int R,int lt)
{
if(!rt||na)return ;
if(L<=l&&r<=R)na=(Mn[rt]<=lt);
else
{
if(L<=Mid)Query(lc[rt],lson,L,R,lt);
if(R>Mid)Query(rc[rt],rson,L,R,lt);
}
}
};
Segment_Tree T;
#undef Mid
#undef lson
#undef rson
int main()
{
while(scanf("%d",&opt)!=EOF&&opt!=3)
{
if(opt==0)T.init();
if(opt==1)
{
int x,y,c;read(x);read(y);read(c);
T.Update(T.root[c],1,N,y,x);
}
if(opt==2)
{
int x,y1,y2,ans=0;read(x);read(y1);read(y2);
if(y1>y2)std::swap(y1,y2);
for(register int i=0;i<=C;++i)na=0,T.Query(T.root[i],1,N,y1,y2,x),ans+=na;
write(ans,'
');
}
}
return 0;
}