牛客暑假多校第二场J-farm

一、题意

White Rabbit has a rectangular farmland of n*m. In each of the grid there is a kind of plant. The plant in the j-th column of the i-th row belongs the a[i][j]-th type.
White Cloud wants to help White Rabbit fertilize plants, but the i-th plant can only adapt to the i-th fertilizer. If the j-th fertilizer is applied to the i-th plant (i!=j), the plant will immediately die.
Now White Cloud plans to apply fertilizers T times. In the i-th plan, White Cloud will use k[i]-th fertilizer to fertilize all the plants in a rectangle [x1[i]...x2[i]][y1[i]...y2[i]].
White rabbits wants to know how many plants would eventually die if they were to be fertilized according to the expected schedule of White Cloud.

简单的说就是，给定一个大型矩形，矩形中每个格子对应一个单独的数字。之后K个命令给，要求大矩阵中某些子矩形的各个元素的赋予统一值，求在命令结束后还保持原有值的个数。其中矩阵的格子总数不超过1e6，但是不指定长宽。

二、解题思路

要求使用一种数据结构进行批量染色操作，且要求后面可以检测是否被"其他颜色污染"。

则看了题解很容易想到，使用某种数据结构做批量增加的操作，之后检测增加后的值是否能够整除原来的元素。

考虑如果直接按照1、2、3、4进行赋值就会有：同样是染色2次，有3+3 = 6和2+2+2 = 6，无法有效判断整除。考虑加一个操作：增加染色次数的判定，但是同样也会有3+3 = 6 = 2+4的问题，因此对数组进行重新设计，则直觉告诉我们，1，2，4，7，......an,an+n这个数列可以完美解决这个问题——不存在第2种组合可以使用相同数目的数列元素相加得到数列的某个其他元素。

因此，使用一位数组开足够大的数组之后，动态的按照二维数组的方式进行寻址，即可完成上述操作。

#include<bits/stdc++.h>
using namespace std;

#define ll intmax_t

const int MAXN=;

ll mapp[MAXN];
ll farm[MAXN];
ll times[MAXN];
ll color[MAXN];

int maxx_numebr =  ;
int add_number =  ;

int m,n,k;

void insert_mex(ll *v,int a,int b,ll key)
{
    a+=;
    b+=;
    while(a<n+)
    {
        int num = a*(m+);
        int pos = b;
        while(pos<m+)
        {
            v[num+pos] += key;
            pos+= pos&(-pos);
        }a+=a&(-a);

    }
}
ll find_mex(ll *v,int a,int b)
{
    a+=;
    b+=;
    ll cntt = ;
    while(a)
    {
        int num = a*(m+);
        int pos = b;
        while(pos)
        {
            cntt += v[num+pos];
            pos -= pos&(-pos);
        }
        a-=a&(-a);
    }
    return cntt;
}

void insert(ll *v,int a,int b,int c,int d,ll key)
{
    // cout<<"coor :"<<a<<" "<<b<<endl;
    // cout<<"coor :"<<a<<" "<<d<<endl;
    // cout<<"coor :"<<c<<" "<<b<<endl;
    // cout<<"coor :"<<c<<" "<<d<<endl;

    insert_mex(v,a,b,key);
    insert_mex(v,c,d,key);
    insert_mex(v,a,d,-key);
    insert_mex(v,c,b,-key);
}

ll find(ll *v,int a,int b)
{
    ll cntt = ;
    cntt += find_mex(v,a,b);
    return cntt;
}

void init()
{
    memset(color,-,sizeof(color));
    for(int i=;i<n;++i)
    {
        int pos = i*m;
        for(int j=;j<m;++j)
        {
            // int ppos = pos +j;
            scanf("%d",&mapp[pos]);
            if(color[pos] == -)color[mapp[pos]] = (maxx_numebr += (add_number++));
            pos++;
        }
    }
    for(int i=;i<k;++i)
    {
        int a,b,c,d,kk;
        scanf("%d%d%d%d%d",&a,&b,&c,&d,&kk);
        if(color[kk] == -)color[kk] = (maxx_numebr += add_number++ );
        ll key = color[kk];
        a--;b--;
        insert(farm,a,b,c,d,key);
        insert(times,a,b,c,d,);
    }
    int cntt = n*m;
    int pos = ;
    for(int i=;i<n;++i)
    {
        for(int j=;j<m;++j)
        {
            // int pos = i*(m+23)+j;
            ll kk = color[mapp[pos]];
            int time_now = find(times,i,j);
            ll res = find(farm,i,j);
            // cout<<"check: "<<i<<" "<<j<<" times: "<<time_now<<" color "<<mapp[pos]<<endl;
            if(time_now ==  || res == time_now*kk)cntt--;
            pos++;
        }
    }
    cout<<cntt<<endl;

}

int main()
{
    cin>>n>>m>>k;
    init();

    return ;
}