Codeforces Round #689 (Div. 2, based on Zed Code Competition)B. Find the Spruce(dp + 重复利用)

B. Find the Spruce

题意

给你一个n行m列由*或者.组成的图形，一个图形高度为k时当且仅当满足\(1 \leq i \leq k\)时，第\(x + i - 1\)行必须满足在区间\([y - i + 1, y +i-1]\)全部都是'*'

思路

暴力解决方法

我们暴力判断当前行是否满足，如果满足我们继续将k加大然后继续向下判断，判断k的整个过程具有连续性，比如说当\(k=3\)时会重复判断\(k=2\)时的情况，所以直接继续判断即可。如果不满足则跳出。

#include<bits/stdc++.h>
using namespace std;
const int N = 2e5 + 100;
typedef long long LL;
//#define int long long 
 
char str[505][505];
int sum[505][505];
int n, m;

void solve() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; ++i) {
		getchar();
		for (int j = 1; j <= m; ++j) {
			scanf("%c", &str[i][j]);
			if (str[i][j] == '*') {
				sum[i][j] = sum[i][j - 1]  + 1;
			}else sum[i][j] = sum[i][j - 1];
		}
	}
	LL ans = 0;
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= m; ++j) {
			for (int k = 1;; ++k) {
				  int l = i + k - 1;
				  int l1 = j - k + 1, r1 = j + k - 1;
				  if (l > n || l1 <= 0 || r1 > m) break;
				  if (sum[l][r1] - sum[l][l1 - 1] != 2 * k - 1) break;
				  ++ans;
			}
		}
	}
	printf("%lld\n", ans);
}
signed main() {
	int T = 1;
	scanf("%d", &T);
	// cin >> T;
	while (T--) {
		solve();
	}
 
}

从下往上dp

如果当前point(i,j)是一个满足条件的图形，当且仅当\(str[i + 1][j - 1],str[i + 1][j], str[i+1][j+1] = '*'\)那么我们从下往上dp整个过程，注意取\(min\)

#include<bits/stdc++.h>
using namespace std;
const int N = 2e5 + 100;
typedef long long LL;
//#define int long long 
 
char str[505][505];
int dp[505][505];
int n, m;


void solve() {
	scanf("%d%d", &n, &m);
	memset(dp, 0, sizeof(dp));
	for (int i = 1; i <= n; ++i) {
		getchar();
		for (int j = 1; j <= m; ++j) {
			scanf("%c", &str[i][j]);
			if (str[i][j] == '*') {
				dp[i][j] = 1;
			}
		}
	}
	LL ans = 0;
	for (int i = n; i  >= 1; --i) {
		for (int j = 1; j <= m; ++j) {
			if(str[i][j] == '*') {
				dp[i][j] = min({dp[i + 1][j - 1], dp[i + 1][j], dp[i + 1][ j + 1]}) + 1;
				ans += dp[i][j];
			}
		}
	}
	printf("%lld\n", ans);

}
signed main() {

	int T = 1;
	scanf("%d", &T);
	// cin >> T;
	while (T--) {
		solve();
	}
 
}