farm
链接：https://www.nowcoder.com/acm/contest/140/J
来源：牛客网
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.
示例1
输入
2 2 2 1 2 2 3 1 1 2 2 2 2 1 2 1 1
输出
3
题目大意： 给出一个矩阵，矩阵里每一个格子里边有一个数字，代表这朵花可以接受的肥料编号，如果浇了别的编号的肥料，这朵花就会死亡。
解题思路： 题解是把1e6内的数字都用一个随机数替代。然后每次对二维区间修改的时候，就用这个随机数去更新。如果只浇了一种肥料，那么该点的值%该点原来的值一定等于0.
#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<vector>
#include<ctime>
#include<cmath>
#include<algorithm>
using namespace std;
#define LL long long
#define inf 0x3f3f3f3f
#define lowb(x) x&-x
#define rep(i,j,k) for(int i=j;i<=k;i++)
#define per(i,j,k) for(int i=j;i>=k;i--)
#define N 10005
#define pb(x) push_back(x)
#define esp 1e-6
#define debug cout<<"------debug------"<<endl;
vector<LL>mp[1000005];
vector<LL>sum[1000005];
LL h[1000005];
int n,m,t;
void init()
{
for(int i=1;i<=1000000;i++)
{
h[i]=i*i+1e9+7;
}
}
void add(int x,int y,LL k)
{
for(int i=x;i<=n;i+=lowb(i))
{
for(int j=y;j<=m;j+=lowb(j))
{
sum[i][j]+=k;
}
}
}
LL query(int x,int y)
{
LL ans=0;
for(int i=x;i>0;i-=lowb(i))
{
for(int j=y;j>0;j-=lowb(j))
{
ans+=sum[i][j];
}
}
return ans;
}
int main()
{
scanf("%lld%lld%lld",&n,&m,&t);
init();
int tem;
rep(i,1,n)
{
mp[i].resize(m+1);
sum[i].resize(m+1);
sum[i][0]=0;
rep(j,1,m)
{
scanf("%d",&tem);
mp[i][j]=h[tem];
}
}
int x1,y1,x2,y2,w;
rep(i,1,t)
{
scanf("%d%d%d%d%d",&x1,&y1,&x2,&y2,&w);
add(x1,y1,h[w]);
add(x2+1,y2+1,h[w]);
add(x2+1,y1,-h[w]);
add(x1,y2+1,-h[w]);
}
int ss=0;
rep(i,1,n)
{
rep(j,1,m)
{
if(query(i,j)%mp[i][j]!=0)ss++;
}
}
printf("%d\n",ss);
}