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

1 1000000 1000000 50

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

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 ;

}

巴特西

HDU 6183 Color it cdq分治 + 线段树 + 状态压缩

Color it

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