二维区间前缀和更新与维护

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.

示例1

输入

复制

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

输出

复制

3

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
using namespace std;
using namespace __gnu_cxx;

const int N = 1e6+10;
int n,m,q;
int a[N];
int vis[N];
vector<long long>v[N],cnt[N],sum[N];
int main()
{
    while(scanf("%d%d%d",&n,&m,&q)!=EOF)
    {
        for(int i = 1; i <= n*m; i++)a[i] = i;
        random_shuffle(a+1,a+n*m+1);
        for(int i = 0; i <= n+1; i++)
        {
            v[i].resize(m+5);
            cnt[i].resize(m+5);
            sum[i].resize(m+5);
        }
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= m; j++)
            {
                scanf("%d",&v[i][j]);
                v[i][j] = a[v[i][j]];
                cnt[i][j] = 0;
                sum[i][j] = 0;
            }
        }
        int x1,x2,y1,y2,k;
        while(q--)
        {
            scanf("%d%d%d%d%d",&x1,&y1,&x2,&y2,&k);
            sum[x1][y1] += a[k];
            sum[x1][y2+1]-=a[k];
            sum[x2+1][y1]-=a[k];
            sum[x2+1][y2+1]+=a[k];
            cnt[x1][y1]++;
            cnt[x1][y2+1]--;
            cnt[x2+1][y1]--;
            cnt[x2+1][y2+1]++;
        }
        int answer = 0;
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= m; j++)
            {
                sum[i][j] += sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1];
                cnt[i][j] += cnt[i-1][j]+cnt[i][j-1]-cnt[i-1][j-1];
                if(sum[i][j]!=cnt[i][j]*v[i][j])answer++;
            }
        }
        cout<<answer<<endl;
    }
    return 0;
}