題意
在一個邊長爲n的正方形矩陣中,尋找邊長爲2~n的全1矩陣的個數
題解
按邊長找,每次的g[i][j](bool)表示當前邊長下起點爲i,j的矩陣是否符合要求
若g[i][j]、g[i + 1][j]、g[i][j + 1]、g[i + 1][j + 1]均爲true,則新的g[i][j]也爲true,否則爲false
我自己想出來的O(\(n^3\))算法
調試記錄
無
#include <cstdio>
#include <cstring>
#define maxn 255
using namespace std;
bool pre[maxn][maxn], next[maxn][maxn];
int n, cnt;
int main(){
scanf("%d", &n);
char str[maxn];
for (int i = 1; i <= n; i++){
scanf("%s", str);
for (int j = 0; j < n; j++) pre[i][j + 1] = str[j] - '0';
}
for (int len = 2; len <= n; len++){
cnt = 0;
for (int x = 1; x <= n - len + 1; x++){
for (int y = 1; y <= n - len + 1; y++){
if (pre[x][y] && pre[x][y + 1] && pre[x + 1][y] && pre[x + 1][y + 1]) next[x][y] = true, cnt++;
}
}
if (!cnt) return 0;
for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) pre[i][j] = next[i][j], next[i][j] = 0;
printf("%d %d\n", len, cnt);
}
return 0;
}