牛客多校第二场 J.Farm

题目链接：https://www.nowcoder.com/acm/contest/140/J

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

将每一个农田上的数字看成2进制，如果多次施肥中同一块农田上施肥的种类不同，那么这些数字必有一位二进制数不同。

先统计每一块农田被使了多少次肥。即统计每一块农田，二进制的每一位有多少个0和1。

然后按位遍历，若0，1同时出现，则植物死亡。

#include
#define maxn 1000010
using namespace std;
int a[maxn],x1[maxn],x2[maxn],y2[maxn],num[maxn],S[maxn],T[maxn],ans[maxn];
int y_1[maxn];
int n,m,t;
int id(int x,int y)
{
    return (x-1)*m+y;
}
void add(int i,int *s)//add和work统计施肥次数
{
    s[id(x1[i],y_1[i])]++;
    if(y2[i]+1<=m) s[id(x1[i],y2[i])+1]--;
    if(x2[i]+1<=n) s[id(x2[i]+1,y_1[i])]--;
    if(x2[i]+1<=n&&y2[i]+1<=m)
        s[id(x2[i]+1,y2[i]+1)]++;
}
void work(int *s)
{
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            if(j>1) s[id(i,j)]+=s[id(i,j-1)];
            if(i>1) s[id(i,j)]+=s[id(i-1,j)];
            if(i>1&&j>1) s[id(i,j)]-=s[id(i-1,j-1)];
        }
    }
}

int main()
{
    while(scanf("%d%d%d",&n,&m,&t)!=EOF)
    {
        memset(ans,0,sizeof(ans));
        for(int i=1;i<=n*m;i++)
            scanf("%d",&a[i]);
        for(int i=1;i<=t;i++)
        {
            scanf("%d%d%d%d%d",&x1[i],&y_1[i],&x2[i],&y2[i],&num[i]);
            add(i,S);
        }
        work(S);
        for(int u=0;u<20;u++)//按位遍历
        {
            memset(T,0,sizeof(T));
            for(int i=1;i<=t;i++)
            {
                if((num[i]>>u)&1)//只统计1出现的次数
                    add(i,T);
            }
            work(T);
            for(int i=1;i<=n*m;i++)
            {
                bool flag;
                if((a[i]>>u)&1)//若植物在这一位上的数是1
                    flag=true;
                else
                    flag=false;
                if(flag)//这一位是1的情况
                {
                    if(S[i]>T[i])//如果总数比1的个数多说明有0，则植物死
                        ans[i]=1;
                }
                else//这一位是0的情况
                {
                    if(T[i])//若出现1，则植物死
                        ans[i]=1;
                }
            }
        }
        int k=0;
        for(int i=1;i<=n*m;i++)
        {
            if(ans[i])
                k++;
        }
        printf("%d\n",k);
    }
    return 0;
}