NC53676「土」秘法地震

题目地址:

https://ac.nowcoder.com/acm/problem/53676

基本思路:

非常裸的一道二维前缀和,我们做出这个矩阵的二维前缀和,每次判断是否会停止施法就是了。关于二维前缀和的求法为递推式:,我们可以大脑中想象一下,应该很容易想到这个递推式是合理的,然后二维前缀和的查询方式,假设我们要找,,到,这部分子矩阵的和那么答案就是 这个大家可以自己画图辅助理解。(我拿鼠标画图疯狂手抖,就不画了OAO)

参考代码:

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
using namespace std;
#define IO std::ios::sync_with_stdio(false)
#define ll long long
#define rep(i, l, r) for (int i = l; i <= r; i++)
#define per(i, l, r) for (int i = l; i >= r; i--)
#define mset(s, _) memset(s, _, sizeof(s))
#define pb push_back
#define pii pair <int, int>
#define mp(a, b) make_pair(a, b)
#define INF 0x3f3f3f3f

inline int read() {
  int x = 0, neg = 1; char op = getchar();
  while (!isdigit(op)) { if (op == '-') neg = -1; op = getchar(); }
  while (isdigit(op)) { x = 10 * x + op - '0'; op = getchar(); }
  return neg * x;
}
inline void print(int x) {
  if (x < 0) { putchar('-'); x = -x; }
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}

const int maxn = 1010;
char c;
int n,m,k,sum[maxn][maxn];
signed main() {
  IO;
  cin >> n >> m >> k;
  mset(sum,0);
  rep(i,1,n){
    rep(j,1,m) cin >> c,sum[i][j] = (c - '0');
  }
  rep(i,1,n){
    rep(j,1,m) sum[i][j] += sum[i-1][j] + sum[i][j-1] - sum[i-1][j-1];
  }
  int ans = 0;
  rep(i,k,n){
    rep(j,k,m){
      int res = sum[i][j] - sum[j-k][j] - sum[i][j-k] + sum[i-k][j-k];
      if(res > 0) ans++;
    }
  }
  cout << ans << '\n';
  return 0;
}