Ultraviolet A 10593-Kites (DP)

Link to the question: Ultraviolet A 10593-Kites

A map of N * n is given, indicating a piece of cart and asking how many methods can be used to make a kite. The kite must be a square or a diamond and cannot have holes.

Solution: the solution is divided into square and diamond:

• Square, DP [I] [J] indicates a square in the lower right corner with I and j
  DP [I] [J] = min (DP [I? 1] [J], DP [I] [J? 1])
  And if the yellow part is 'x', DP [I] [J] ++
  
  

• Diamond, DP [I] [J] indicates the positive direction of the Diamond
  
  If the yellow part of the city is 'x', DP [I] [J] ++

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 505;int n, dp[N][N];char g[N][N];int solve () {    int ans = 0;    memset(dp, 0, sizeof(dp));    for (int i = 1; i <= n; i++) {        for (int j = 1; j <= n; j++) {            if (g[i][j] == ‘x‘) {                int tmp = min(dp[i-1][j], dp[i][j-1]);                dp[i][j] = tmp + (g[i-tmp][j-tmp] == ‘x‘);                if (dp[i][j] > 1)                    ans += dp[i][j] - 1;            }        }    }    memset(dp, 0, sizeof(dp));    for (int i = 1; i <= n; i++) {        for (int j = 1; j <= n; j++) {            if (g[i][j] == ‘x‘) {                int tmp = min(dp[i-1][j-1], dp[i-1][j+1]);                if (tmp == 0 || g[i-1][j] != ‘x‘)                    dp[i][j] = 1;                else if (g[i-tmp*2][j] == ‘x‘ && g[i-tmp*2+1][j] == ‘x‘)                    dp[i][j] = tmp + 1;                else                    dp[i][j] = tmp;                if (dp[i][j] > 1)                    ans += dp[i][j] - 1;            }        }    }    return ans;}int main () {    while (scanf("%d%*c", &n) == 1 && n) {        for (int i = 1; i <= n; i++)            gets(g[i]+1);        printf("%d\n", solve());    }    return 0;}