# (BFS 图的遍历) leetcode 1020. Number of Enclaves

Given a 2D array A, each cell is 0 (representing sea) or 1 (representing land)

A move consists of walking from one land square 4-directionally to another land square, or off the boundary of the grid.

Return the number of land squares in the grid for which we cannot walk off the boundary of the grid in any number of moves.

Example 1:

Input: [[0,0,0,0],[1,0,1,0],[0,1,1,0],[0,0,0,0]]
Output: 3
Explanation:
There are three 1s that are enclosed by 0s, and one 1 that isn't enclosed because its on the boundary.

Example 2:

Input: [[0,1,1,0],[0,0,1,0],[0,0,1,0],[0,0,0,0]]
Output: 0
Explanation:
All 1s are either on the boundary or can reach the boundary.


Note:

1. 1 <= A.length <= 500
2. 1 <= A[i].length <= 500
3. 0 <= A[i][j] <= 1
4. All rows have the same size.

C++代码：

int dx[] = {0,0,1,-1};
int dy[] = {1,-1,0,0};
class Solution {
public:
typedef pair<int,int> pii;
queue<pii> q;
int numEnclaves(vector<vector<int>>& A) {
int m = A.size();
int n = A[0].size();
for(int i = 0; i < m; i++){
if(A[i][0] == 1)
A[i][0] = 2;
if(A[i][n-1] == 1)
A[i][n-1] = 2;
}
for(int j = 0; j < n; j++){
if(A[0][j] == 1){
A[0][j] = 2;
}
if(A[m-1][j] == 1){
A[m-1][j] = 2;
}
}
for(int i = 0; i < m; i++){
for(int j = 0; j < n; j++){
if(A[i][j] == 2){
q.push(make_pair(i,j));
while(!q.empty()){
auto t = q.front();q.pop();
for(int i = 0;i < 4; i++){
int x = t.first + dx[i];
int y = t.second + dy[i];
if(x >= 0 && x < m && y >= 0 && y < n && A[x][y] == 1){
q.push(make_pair(x,y));
A[x][y] = 2;
}
}
}
}
}
}
int cnt = 0;
for(int i = 0; i < m; i++){
for(int j = 0; j < n; j++){
if(A[i][j] == 1)
cnt++;
}
}
return cnt;
}
};

posted @ 2019-05-15 21:19  PJCK  阅读(36)  评论(0编辑  收藏