HDU 6183 Color it cdq分治 + 线段树 + 状态压缩

Color it

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others)

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

 

题解：

 

　　对于加入的点，我把第一维， x 进行排序， cdq分治优化时间这一维，其余部分用线段树

　　因为只有50中颜色，我将每一位颜色进行二进制压缩，当作一个数存在线段树里

　　利用线段树查询一个区间有多少不同的颜色（位运算）

　　

#include <bits/stdc++.h>

inline int read(){int x=,f=;char ch=getchar();while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}return x*f;}
using namespace std;
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
const int N = 1e6 + ;

namespace IO {
    const int MX = 4e7; //1e7占用内存11000kb
    char buf[MX]; int c, sz;
    void begin() {
        c = ;
        sz = fread(buf, , MX, stdin);
    }
    inline bool read(int &t) {
        while(c < sz && buf[c] != '-' && (buf[c] < '' || buf[c] > '')) c++;
        if(c >= sz) return false;
        bool flag = ; if(buf[c] == '-') flag = , c++;
        for(t = ; c < sz && '' <= buf[c] && buf[c] <= ''; c++) t = t *  + buf[c] - '';
        if(flag) t = -t;
        return true;
    }
}

struct ss{
    int op,x,y,z,id;
    long long ans;
    ss(int op = ,int x = ,int y = ,int z = ,int id = ,long long ans = ) : op(op), x(x), y(y), z(z), id(id), ans(ans) {}
}Q[N],t[N];

bool cmp(ss s1,ss s2) { if(s1.x == s2.x) return s1.op < s2.op;else return s1.x < s2.x; }

int n,mx,san[N];
void init() {n = ;}
long long v[N * ];

void update(int i,int ll,int rr,int x,long long c,int ff) {
    if(ll == rr && x == ll) {if(!ff)v[i] |= 1LL<<c;else v[i] = c;return ;}
    if(x <= mid) update(ls,ll,mid,x,c,ff);
    else update(rs,mid+,rr,x,c,ff);
    v[i] = v[ls] | v[rs];
}
long long ask(int i,int ll,int rr,int x,int y) {
    if(x > y) return ;
    if(ll == x && rr == y) return v[i];
    if(y <= mid) return ask(ls,ll,mid,x,y);
    else if(x > mid) return ask(rs,mid+,rr,x,y);
    else return (ask(ls,ll,mid,x,mid) | ask(rs,mid+,rr,mid+,y));
}
void cdq(int ll,int rr) {
    if(ll == rr) return ;
    for(int i = ll; i <= rr; ++i) {
        if(Q[i].id <= mid && Q[i].op == )
            update(,,mx,Q[i].y,Q[i].z,);
        else if(Q[i].id > mid && Q[i].op == )
            Q[i].ans |= ask(,,mx,Q[i].y,Q[i].z);
    }
    for(int i = ll; i <= rr; ++i) {
        if(Q[i].id <= mid && Q[i].op == )
            update(,,mx,Q[i].y,,);
    }
    int L1 = ll, R1 = mid+;
    for(int i = ll; i <= rr; ++i) {
        if(Q[i].id <= mid) t[L1++] = Q[i];
        else t[R1++] = Q[i];
    }
    for(int i = ll; i <= rr; ++i) Q[i] = t[i];
    cdq(ll,mid);cdq(mid+,rr);
}
void solve() {
    if(n == ) return ;
    int cny = ;
    for(int i = ; i <= n; ++i) {
        if(Q[i].op == ) san[++cny] = Q[i].y;
        else {
            san[++cny] = Q[i].y;
            san[++cny] = Q[i].z;
        }
    }
    sort(san+,san+cny+);
    int SA = unique(san+,san+cny+) - san - ;
    for(int i = ; i <= n; ++i) {
        if(Q[i].op == ) Q[i].y = lower_bound(san+,san+SA+,Q[i].y) - san;
        else {
            Q[i].y = lower_bound(san+,san+SA+,Q[i].y) - san;
            Q[i].z = lower_bound(san+,san+SA+,Q[i].z) - san;
        }
    }
    mx = SA;
    sort(Q+,Q+n+,cmp);
    cdq(,n);
    for(int i = ; i <= n; ++i) {
        if(Q[i].op == ) {
            int sum = ;
            for(int j = ; j <= ; ++j)
                if((Q[i].ans >> j) & ) sum++;
            printf("%d\n",sum);
        }
    }
}
int op,x,z,y;
int main() {
    IO::begin();
    while() {
        IO::read(op);
        if(op ==  || op == ) {
            solve();
            init();
            if(op == ) return ;
            continue;
        }
        IO::read(x);
        IO::read(y);
        IO::read(z);
        Q[++n] = ss(op,x,y,z,n,);
    }
    return ;
}