farm

farm

时间限制：C/C++ 4秒，其他语言8秒
空间限制：C/C++ 262144K，其他语言524288K
64bit IO Format: %lld

题目描述

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.

输入描述:

The first line of input contains 3 integers n,m,T(n*m<=1000000,T<=1000000)
For the next n lines, each line contains m integers in range[1,n*m] denoting the type of plant in each grid.
For the next T lines, the i-th line contains 5 integers x1,y1,x2,y2,k(1<=x1<=x2<=n,1<=y1<=y2<=m,1<=k<=n*m)

输出描述:

Print an integer, denoting the number of plants which would die.

输入

2 2 2
1 2
2 3
1 1 2 2 2
2 1 2 1 1

输出

3

---

题意：n*m的矩阵里，每个格子都有一个数，有t次操作，每次都会选一个小矩形区域并且选择一个数x，这个小矩形区域里的数要是不等于x，这个数就会消失，问最后有多少个数会消失。
反过来想，这个数要是不消失，那么一旦这个数被选中，x肯定是等于这个数的。若把每一次的操作变成给矩形区域里都加上数x，那么最后若一个格子里的总和可以整除格子里的数，这个数则会有很大概率不消失。
于是考虑给矩阵里的数重新分配权值，且每次操作的x值也变为新的权值，这样就会很大概率避免错误情况。

#include<bits/stdc++.h>
#define N 1000100
using namespace std;
 
vector<int>Map[N];
vector<long long>dis[N];
int Rand[N];
 
int main()
{
    int n,m,t;
    scanf("%d %d %d",&n,&m,&t);
 
    for(int i=1;i<=n;i++)
    {
        Map[i].push_back(0);
        dis[i].push_back(0);
       for(int j=1;j<=m;j++)
       {
            int a;
            scanf("%d",&a);
            Map[i].push_back(a);
            dis[i].push_back(0);
        }
    }
    srand(time(0));
    for(int i=1;i<=n*m;i++)Rand[i]=rand()*rand()%9000000+rand();
 
    while(t--)
    {
        int a,b,c,d,k;
        scanf("%d %d %d %d %d",&a,&b,&c,&d,&k);
 
        dis[a][b]+=Rand[k];
        if(c+1<=n) dis[c+1][b]-=Rand[k];
        if(d+1<=m)dis[a][d+1]-=Rand[k];
        if(c+1<=n&&d+1<=m) dis[c+1][d+1]+=Rand[k];
 
    }
 
    for(int i=1;i<=n;i++)
    {
        long long now=0;
        for(int j=1;j<=m;j++)
        {
            now+=dis[i][j];
            if(i-1>=1)
            dis[i][j]=dis[i-1][j]+now;
            else
                dis[i][j]=now;
        }
    }
 
    int ans=0;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
            if(dis[i][j]%Rand[Map[i][j]]!=0)ans++;
 
    printf("%d",ans);
    return 0;
 
}