farm

链接：https://www.nowcoder.com/acm/contest/140/J
来源：牛客网
 

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

题意：

在n*m的矩形里，给植物施肥，如果不匹配植物就会死去，问植物死了多少

分析：

如果用树状数组更新区间，再还原过85%，就会TLE，所以用判断该点改变的总和跟（改变的次数与该点的数的乘积）是否相等，这样做会wrong，不会TLE，wrong是因为存在当原数为3，更新4次，分别为1、5、2、4这种情况下，和恰好为改变的次数与该点的数的乘积，这个时候可以做一个映射，消除数与数之间的联系（可以用随机数实现）

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<stdlib.h>
using namespace std;
typedef long long ll;
const int maxn=1e6+10;
int a[maxn],cnt[maxn],ran[maxn];
ll sum[maxn];
int n,m,t,x,y,z,w,k;
int lowbit(int x){
    return x&-x;
}
void update(int x,int y,int add,int d)
{
    int tempy=y;
    while(x<=n)
    {
        y=tempy;
        while(y<=m)
        {
            sum[(x-1)*m+y-1]+=add;
            cnt[(x-1)*m+y-1]+=d;
            y+=lowbit(y);
        }
        x+=lowbit(x);
    }
}
bool query(int x,int y)
{
    ll rets=0,retc=0;
    int tmp=ran[a[(x-1)*m+y-1]];
    int tempy=y;
    while(x>0){
        y=tempy;
        while(y>0){
            rets+=sum[(x-1)*m+y-1];
            retc+=cnt[(x-1)*m+y-1];
            y-=lowbit(y);
        }
        x-=lowbit(x);
    }
    if(rets==retc*tmp) return true;
    else return false;
}
void init(){
    for(int i=0;i<maxn;i++){
        ran[i]=rand();
    }
}
int main(){
    init();
    while(scanf("%d%d%d",&n,&m,&t)==3){
        memset(cnt,0,sizeof(cnt));
        memset(sum,0,sizeof(sum));
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                scanf("%d",&a[i*m+j]);
            }
        }
        for(int i=0;i<t;i++){
            scanf("%d%d%d%d%d",&x,&y,&z,&w,&k);
            update(z+1,w+1,ran[k],1);
            update(x,y,ran[k],1);
            update(x,w+1,-ran[k],-1);
            update(z+1,y,-ran[k],-1);
        }
        int ans=n*m;
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++)
                if(query(i,j)) ans--;
        }
        printf("%d\n",ans);
    }
    return 0;
}