class Solution {
    public int minPathCost(int[][] grid, int[][] moveCost) {
    int n = grid.length, m = grid[0].length;
    int[][] dp = new int[n][m];//dp[i][j]表示以gril[i][j]结尾的路径的最小值
    int ans = Integer.MAX_VALUE;
    for (int i = 0; i < dp.length; i++) {
      Arrays.fill(dp[i], Integer.MAX_VALUE);
    }
    for (int j = 0; j < m; j++) {
      dp[0][j] = grid[0][j];
    }
    for (int i = 1; i < n; i++) {
      for (int j = 0; j < m; j++) {
        for (int k = 0; k < m; k++) {
          /*
           * dp[i - 1][k] 从dp[i-1][k]到dp[i][j]
           * moveCost[grid[i - 1][k]][j] 从dp[i-1][k]到dp[i][j]的路径的值
           * grid[i][j]  该点的值
           */
          dp[i][j] = Math.min(dp[i - 1][k] + moveCost[grid[i - 1][k]][j] + grid[i][j],
             dp[i][j]);
        }
      }
    }
    n--;//为了方便枚举终点的路径最小值
    for (int j = 0; j < m; j++) {
      ans = Math.min(ans, dp[n][j]);//寻找达到尾部的最小值
    }
    return ans;
  }
}