Title Description
A ground grid and m rows n columns. A robot moves from the grid coordinates 0,0, every time only left, right, upper, lower four directions a cell, but can not enter the row and column coordinates of the grid is greater than the sum of the number of bits k. For example, when k is 18, the robot can enter the box (35, 37), because 3 + 3 + 5 + 7 = 18. However, it can not enter the box (35, 38), because 3 + 3 + 5 + 8 = 19. Will the robot be able to reach the number of lattice?
Problem-solving ideas
动态规划 dp[i][j]表示是否可以到达，统计数字中
true
的个数，即为可以到达的格子数
code show as below
public int movingCount(int threshold, int rows, int cols)
{
if(threshold<0)
return 0;
boolean [][] dp=new boolean[rows+1][cols+1];
dp[0][0]=true;
for(int i=1;i<=rows;i++){//初始化
if(dp[i-1][0]&&canreach(threshold,i,0)){
dp[i][0]=true;
}else {
dp[i][0]=false;
}
}
for(int i=1;i<=cols;i++){//初始化
if(dp[0][i-1]&&canreach(threshold,0,i)){
dp[0][i]=true;
}else {
dp[0][i]=false;
}
}
for(int i=1;i<=rows;i++){
for(int j=1;j<=cols;j++){
if((dp[i-1][j]&&canreach(threshold, i, j))||(dp[i][j-1]&&canreach(threshold, i, j))){
dp[i][j]=true;
}else {
dp[i][j]=false;
}
}
}
int count=0;
for(int i=0;i<rows;i++){
for(int j=0;j<cols;j++){
if(dp[i][j])
count++;
}
}
return count;
}
public boolean canreach(int threshold, int x, int y) {
int sum = 0;
while (x > 0) {
sum += x % 10;
x /= 10;
}
while (y > 0) {
sum += y % 10;
y /= 10;
}
return sum <= threshold;
}