# [LeetCode] 778. Swim in Rising Water 在上升的水中游泳

On an N x N grid, each square grid[i][j] represents the elevation at that point (i,j).

Now rain starts to fall. At time t, the depth of the water everywhere is t. You can swim from a square to another 4-directionally adjacent square if and only if the elevation of both squares individually are at most t. You can swim infinite distance in zero time. Of course, you must stay within the boundaries of the grid during your swim.

You start at the top left square (0, 0). What is the least time until you can reach the bottom right square (N-1, N-1)?

Example 1:

Input: [[0,2],[1,3]]
Output: 3
Explanation:
At time 0, you are in grid location (0, 0).
You cannot go anywhere else because 4-directionally adjacent neighbors have a higher elevation than t = 0.

You cannot reach point (1, 1) until time 3.
When the depth of water is 3, we can swim anywhere inside the grid.


Example 2:

Input: [[0,1,2,3,4],[24,23,22,21,5],[12,13,14,15,16],[11,17,18,19,20],[10,9,8,7,6]]
Output: 16
Explanation:
0  1  2  3  4
24 23 22 21  5
12 13 14 15 16
11 17 18 19 20
10  9  8  7  6

The final route is marked in bold.
We need to wait until time 16 so that (0, 0) and (4, 4) are connected.


Note:

1. 2 <= N <= 50.
2. grid[i][j] is a permutation of [0, ..., N*N - 1].

class Solution {
public:
int swimInWater(vector<vector<int>>& grid) {
int res = 0, n = grid.size();
unordered_set<int> visited{0};
vector<vector<int>> dirs{{0, -1}, {-1, 0}, {0, 1}, {1, 0}};
auto cmp = [](pair<int, int>& a, pair<int, int>& b) {return a.first > b.first;};
priority_queue<pair<int, int>, vector<pair<int, int>>, decltype(cmp) > q(cmp);
q.push({grid[0][0], 0});
while (!q.empty()) {
int i = q.top().second / n, j = q.top().second % n; q.pop();
res = max(res, grid[i][j]);
if (i == n - 1 && j == n - 1) return res;
for (auto dir : dirs) {
int x = i + dir[0], y = j + dir[1];
if (x < 0 || x >= n || y < 0 || y >= n || visited.count(x * n + y)) continue;
visited.insert(x * n + y);
q.push({grid[x][y], x * n + y});
}
}
return res;
}
};

// Time Limit Exceeded (TLE)
class Solution {
public:
vector<vector<int>> dirs{{0, -1}, {-1, 0}, {0, 1}, {1, 0}};
int swimInWater(vector<vector<int>>& grid) {
int n = grid.size();
vector<vector<int>> dp(n, vector<int>(n, INT_MAX));
helper(grid, 0, 0, grid[0][0], dp);
return dp[n - 1][n - 1];
}
void helper(vector<vector<int>>& grid, int x, int y, int cur, vector<vector<int>>& dp) {
int n = grid.size();
if (x < 0 || x >= n || y < 0 || y >= n || max(cur, grid[x][y]) >= dp[x][y]) return;
dp[x][y] = max(cur, grid[x][y]);
for (auto dir : dirs) {
helper(grid, x + dir[0], y + dir[1], dp[x][y], dp);
}
}
};

class Solution {
public:
int swimInWater(vector<vector<int>>& grid) {
int n = grid.size();
int left = grid[0][0], right = n * n;
while (left < right) {
int mid = left + (right - left) / 2;
if (!helper(grid, mid)) left = mid + 1;
else right = mid;
}
return left;
}
bool helper(vector<vector<int>>& grid, int mid) {
int n = grid.size();
unordered_set<int> visited{0};
vector<vector<int>> dirs{{0, -1}, {-1, 0}, {0, 1}, {1, 0}};
stack<int> st{{0}};
while (!st.empty()) {
int i = st.top() / n, j = st.top() % n; st.pop();
if (i == n - 1 && j == n - 1) return true;
for (auto dir : dirs) {
int x = i + dir[0], y = j + dir[1];
if (x < 0 || x >= n || y < 0 || y >= n || visited.count(x * n + y) || grid[x][y] > mid) continue;
st.push(x * n + y);
visited.insert(x * n + y);
}
}
return false;
}
};

Github 同步地址：

https://github.com/grandyang/leetcode/issues/778

https://leetcode.com/problems/swim-in-rising-water/

https://leetcode.com/problems/swim-in-rising-water/discuss/113743/JAVA-DP-+-DFS

https://leetcode.com/problems/swim-in-rising-water/discuss/113765/Easy-and-Concise-Solution-using-Binary-Search-PythonC++

https://leetcode.com/problems/swim-in-rising-water/discuss/113758/C++-two-solutions-Binary-Search+DFS-and-Dijkstra+BFS-O(n2logn)-11ms

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posted @ 2018-05-09 23:26  Grandyang  阅读(5811)  评论(2编辑  收藏  举报