時間限制：C/C++ 4秒，其他語言8秒
空間限制：C/C++ 262144K，其他語言524288K
64bit IO Format: %lld

題目描述

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.

輸入描述:

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)

輸出描述:

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

示例1

輸入

複製

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

輸出

複製

3

題意：

N*M的矩形，裡面種植物，每種植物適應的藥物不同，如果植物受到不適應的藥物就會死，現在T次打藥，每次在一個矩形中打藥，問最終死了多少植物？

分析：二位樹狀陣列+隨機化。對每種植物適應的藥性值進行隨機化，然後用二位樹狀陣列來維護T次矩形區域中每種植物所施加的藥性Hash值的總和，最終判斷藥性總和是否為植物所適應的藥性Hash值的倍數。

程式碼：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <string>
#include <vector>
#include <queue>
#include <cmath>
#include <set>
#include <vector>
#include <stack>
#define lowbit(x) x&(-x)
using namespace std;
typedef long long ll;
const int maxn=1e6+10;
vector<ll> G[maxn];
vector<ll> ans[maxn];
ll Hash[maxn];
int n,m,t;
void init()
{
    for(int i=1;i<=1e6+6;i++)
    {
        Hash[i]=rand()*1e6+rand()*rand();
    }
}
void update(int x,int y,ll val)
{
    for(int i=x;i<=n;i+=lowbit(i))
    {
        for(int j=y;j<=m;j+=lowbit(j))
        {
            ans[i][j]+=val;
        }
    }
}
ll qury(int x,int y)
{
    ll res=0;
    for(int i=x;i>=1;i-=lowbit(i))
    {
        for(int j=y;j>=1;j-=lowbit(j))
        {
            res+=ans[i][j];
        }
    }
    return res;
}
int main()
{
    init();
    scanf("%d%d%d",&n,&m,&t);
    for(int i=1;i<=n;i++)
    {
        G[i].push_back(0);
        ans[i].push_back(0);
        for(int j=1;j<=m;j++)
        {
            int x;
            scanf("%d",&x);
            G[i].push_back(Hash[x]);
            ans[i].push_back(0);
        }
    }
    for(int i=1;i<=t;i++)
    {
        int x1,x2,y1,y2,k;
        scanf("%d%d%d%d%d",&x1,&y1,&x2,&y2,&k);
        update(x1,y1,Hash[k]);
        update(x2+1,y2+1,Hash[k]);
        update(x1,y2+1,-Hash[k]);
        update(x2+1,y1,-Hash[k]);
    }
    int sum=0;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            ll s=qury(i,j);
            if(s%G[i][j])
                sum++;
        }
    }
    printf("%d\n",sum);
	return 0;
}