c语言记忆化搜索

说白了，走过的路，不再走

一个点向周围虹吸，这样不用重复算

//记忆化搜索
const int dirs[4][2] = {{-1,0},{1,0},{0,-1},{0,1}};
const int mod = 1e9 + 7;
long long dp(int i ,int j,int** grid,int m,int n,long long** hash)
{
    if(hash[i][j] > 0)
    {
        return hash[i][j];
    }
    hash[i][j] = 1;
    for(int k = 0;k<4;k++)
    {
        int x = i + dirs[k][0],y = j + dirs[k][1];
        if(x < 0 || y<0 || x >= m || y>= n || grid[x][y] >= grid[i][j])continue;
        hash[i][j] = (hash[i][j] + dp(x,y,grid,m,n,hash))%mod;
    }
    return hash[i][j];
}

int countPaths(int** grid, int gridSize, int* gridColSize){
    int m = gridSize;
    int n = gridColSize[0];
    long long** hash = (long long**)malloc(sizeof(long long*)*gridSize);
    for(int i = 0;i<gridSize;i++)
    {
        hash[i] = malloc(sizeof(long long)*gridColSize[0]);
    }
    for(int i =0 ;i<m;i++)
    {
        for(int j = 0;j<n;j++)
        {
            hash[i][j] = 0;
        }
    }
   long long ans = 0;
    for(int i = 0;i<m;i++)
    {
        for(int j = 0;j<n;j++)
        {
            ans = (ans+dp(i,j,grid,gridSize,gridColSize[0],hash))%mod;
        }
    }
    return ans;

}