18牛客第二场 J farm

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

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

题意:给出一个n*m的区域,t次操作 每个点都是一种植物 t次操作区域覆盖肥料，如果区域里面的植物不等于覆盖的飞亮那么就植物死亡，求有多少植物死亡。

思路:我们将每种植物的坐标存下来，然后我也将每种肥料的覆盖区域坐标存下，在我们算每种植物贡献的时候，我们先删除同种肥料对于贡献的影响，然后再查询完再加上

最后枚举所有植物即可。

#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define pll pair<int,int>
#define mp make_pair
#define ll long long
const int maxn = 1000000+5;
struct node
{
    int x1,x2,y1,y2;
    node(int x1,int x2,int y1,int y2):x1(x1),x2(x2),y1(y1),y2(y2) {};
};
vector< vector< pll > > p;
vector< vector< int > > a;
vector< vector<node> > pos;
int n,m;
int lowbit(int x)
{
    return x&(-x);
}
void add(int x,int y,int d)
{

    for(int i=x; i<=n; i+=lowbit(i))
        for(int j=y; j<=m; j+=lowbit(j))
            a[i][j]+=d;
}
int query(int x,int y)
{
    int ans=0;
    for(int i=x; i>0; i-=lowbit(i))
        for(int j=y; j>0; j-=lowbit(j))
            ans+=a[i][j];
    return ans;

}
void updata(int x1,int y1,int x2,int y2,int d)
{
    add(x1,y1,d);
    add(x2+1,y2+1,d);
    add(x1,y2+1,-d);
    add(x2+1,y1,-d);
}
int main()
{
    int t,k,x1,x2,y1,y2;
    scanf("%d %d %d",&n,&m,&t);
    a.resize(n+1);
    for(int i=1; i<=n; i++)
        a[i].resize(m+1);
    p.resize(n*m+1);
    pos.resize(n*m+1);
    for(int i=1; i<=n; i++)
        for(int j=1; j<=m; j++)
        {
            scanf("%d",&k);
            p[k].pb(mp(i,j));
        }
    for(int i=1; i<=t; i++)
    {
        scanf("%d %d %d %d %d",&x1,&y1,&x2,&y2,&k);
        pos[k].pb(node(x1,x2,y1,y2));
        updata(x1,y1,x2,y2,1);
    }
    ll ans=0;
    for(int i=1; i<=n*m; i++)
    {
        int len=pos[i].size();
        for(int j=0; j<len; j++)
        {
            updata(pos[i][j].x1,pos[i][j].y1,pos[i][j].x2,pos[i][j].y2,-1);
        }
        int len1=p[i].size();
        for(int j=0; j<len1; j++)
        {
            if(query(p[i][j].first,p[i][j].second))
                ans++;
        }
        for(int j=0; j<len; j++)
        {
            updata(pos[i][j].x1,pos[i][j].y1,pos[i][j].x2,pos[i][j].y2,1);
        }

    }
   printf("%lld
",ans);
}

PS：摸鱼怪的博客分享，欢迎感谢各路大牛的指点~