Problem Statement
We have a grid with H rows and W columns. At first, all cells were painted white.
Snuke painted N of these cells. The i-th ( 1≤i≤N ) cell he painted is the cell at the ai-th row and bi-th column.
Compute the following:
- For each integer j ( 0≤j≤9 ), how many subrectangles of size 3×3 of the grid contains exactly j black cells, after Snuke painted N cells?
Constraints
- 3≤H≤109
- 3≤W≤109
- 0≤N≤min(105,H×W)
- 1≤ai≤H (1≤i≤N)
- 1≤bi≤W (1≤i≤N)
- (ai,bi)≠(aj,bj) (i≠j)
Input
The input is given from Standard Input in the following format:
H W N a1 b1 : aN bN
Output
Print 10 lines. The (j+1)-th ( 0≤j≤9 ) line should contain the number of the subrectangles of size 3×3 of the grid that contains exactly j black cells.
Sample Input 1
4 5 8 1 1 1 4 1 5 2 3 3 1 3 2 3 4 4 4
Sample Output 1
0 0 0 2 4 0 0 0 0 0
There are six subrectangles of size 3×3. Two of them contain three black cells each, and the remaining four contain four black cells each.
Sample Input 2
10 10 20 1 1 1 4 1 9 2 5 3 10 4 2 4 7 5 9 6 4 6 6 6 7 7 1 7 3 7 7 8 1 8 5 8 10 9 2 10 4 10 9
Sample Output 2
4 26 22 10 2 0 0 0 0 0
Sample Input 3
1000000000 1000000000 0
Sample Output 3
999999996000000004 0 0 0 0 0 0 0 0 0 题意：给定一个高为h，宽为w的矩阵，然后给你n个黑色块的坐标。让你求出所有大小为3*3的矩阵分别包含了多少个黑色块，你只需要输出含有0~9个黑色块的个数的矩阵数量分别是多少。 思路：由于h和w的数量很大，没有办法进行直接标记模拟。、我们思考如下：每一个黑色的方块只会对9个3*3的矩阵有贡献。 看图：
看图可以知道，蓝色圆圈的位置如果是黑色块，可以对以红色点为左上角起点的3*3的区间有贡献。
那么我们对每一个黑色块算出的一共9个的贡献矩阵，全部加入到一个数组中，排序后处理答案即可。
细节见代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d\n",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), ‘\0‘, sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define db(x) cout<<"== [ "<<x<<" ] =="<<endl;
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
inline void getInt(int* p);
const int maxn=1000010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
struct node
{
ll x,y;
}a[maxn];
ll n;
ll h,w;
ll xx[]={-2,-2,-2,-1,-1,-1,0,0,0};
ll yy[]={-2,-1,0,-2,-1,0,-2,-1,0};
ll ans[1000];
ll mod=1e9+7;
int main()
{
//freopen("D:\\common_text\\code_stream\\in.txt","r",stdin);
//freopen("D:\\common_text\\code_stream\\out.txt","w",stdout);
gbtb;
cin>>h>>w>>n;
repd(i,1,n)
{
cin>>a[i].x>>a[i].y;
}
vector<ll> v;
repd(i,1,n)
{
repd(j,0,8)
{
ll x=a[i].x+xx[j];
ll y=a[i].y+yy[j];
if(x>=1&&x+2<=h&&y>=1&&y+2<=w)
{
// cout<<x<<" "<<y<<endl;
ll num=(x)*mod+y;
v.push_back(num);
}
}
}
sort(ALL(v));
v.push_back(-9ll);
ll ww=1ll;
ll ans0=(h-2ll)*(w-2ll);
for(int i=0;i<v.size()-1;i++)
{
// db(v[i]);
if(v[i]==v[i+1])
{
ww++;
}else
{
ans[ww]++;
ww=1ll;
ans0--;
}
}
cout<<ans0<<endl;
repd(i,1,9)
{
cout<<ans[i]<<endl;
}
return 0;
}
inline void getInt(int* p) {
char ch;
do {
ch = getchar();
} while (ch == ‘ ‘ || ch == ‘\n‘);
if (ch == ‘-‘) {
*p = -(getchar() - ‘0‘);
while ((ch = getchar()) >= ‘0‘ && ch <= ‘9‘) {
*p = *p * 10 - ch + ‘0‘;
}
}
else {
*p = ch - ‘0‘;
while ((ch = getchar()) >= ‘0‘ && ch <= ‘9‘) {
*p = *p * 10 + ch - ‘0‘;
}
}
}
すぬけ君の塗り絵 / Snuke's Coloring AtCoder - 2068 （思维，排序，贡献）
原文地址：https://www.cnblogs.com/qieqiemin/p/10713861.html