# 1102. Path With Maximum Minimum Value 得分最高的路径

@TOC

## # 题目描述

Given a matrix of integers A with R rows and C columns, find the maximum score of a path starting at [0,0] and ending at [R-1,C-1].

The score of a path is the minimum value in that path. For example, the value of the path 8 → 4 → 5 → 9 is 4.

A path moves some number of times from one visited cell to any neighbouring unvisited cell in one of the 4 cardinal directions (north, east, west, south).

Example 1:

Input: [[5,4,5],[1,2,6],[7,4,6]]
Output: 4
Explanation:
The path with the maximum score is highlighted in yellow.

Example 2:

Input: [[2,2,1,2,2,2],[1,2,2,2,1,2]]
Output: 2

Example 3:

Input: [[3,4,6,3,4],[0,2,1,1,7],[8,8,3,2,7],[3,2,4,9,8],[4,1,2,0,0],[4,6,5,4,3]]
Output: 3

Note:

1. 1 <= R, C <= 100
2. 0 <= A[i][j] <= 10^9

## # 解题方法

### # 排序+并查集

1. 题目的意思是路径上最小的点，并不是最短路径，所以按照值的大小进行排序。
2. 已经完成了按值的排序，所以会把图里面值最大的点优先访问。
3. 每次新遍历一个点的时候，检查周围的点是否已经访问过（值更大），把该点放入周围的图中。
4. 值最大的点不一定在一起，因此会形成多个子图。
5. 直到添加了一个较小的点时，起终两点联通了，那么这个新添加的点就是我们要求的。

C++代码如下：

class Solution {
public:
int maximumMinimumPath(vector<vector<int>>& A) {
const int M = A.size();
const int N = A[0].size();
const int X = M * N;
parent = vector<int>(X, 0);
for (int i = 0; i < parent.size(); ++i)
parent[i] = i;
vector<vector<int>> values;
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
values.push_back({A[i][j], i, j});
}
}
sort(values.begin(), values.end(), [](vector<int>& a, vector<int>& b) {return a[0] < b[0];});
unordered_set<int> visited;
visited.insert(0);
visited.insert(X - 1);
int res = min(A[0][0], A[M - 1][N - 1]);
while(find(0) != find(X - 1)) {
vector<int> cur = values.back(); values.pop_back();
visited.insert(cur[1] * N + cur[2]);
res = min(res, cur[0]);
for (auto& dir : dirs) {
int newx = cur[1] + dir[0];
int newy = cur[2] + dir[1];
if (newx >= 0 && newx < M && newy >= 0 && newy < N && visited.count(newx * N + newy)) {
uni(cur[1] * N + cur[2], newx * N + newy);
}
}
}
return res;
}
int find(int a) {
if (parent[a] == a)
return a;
return find(parent[a]);
}
void uni(int a, int b) {
int pa = find(a);
int pb = find(b);
if (pa == pb)
return;
parent[pa] = pb;
}
private:
vector<int> parent;
vector<vector<int>> dirs = {{0, 1}, {0, -1}, {-1, 0}, {1, 0}};
};

### # 优先级队列

[[6, 6, 0],
[3, 0, 0],
[3, 3, 3]]

起始时把[0,1]位置的6和[1,0]位置的3都放入堆中。
第一次会选择[0,1]位置的6，
但是由于其周围的全是0，0放入堆中排到了后面，
所以第二次访问的是[1,0]位置的3.

C++代码如下：

class Solution {
public:
int maximumMinimumPath(vector<vector<int>>& A) {
int R = A.size();
int C = A[0].size();
vector<vector<int> > visited(R, vector<int>(C, false));
visited[0][0] = true;
priority_queue<Point> pq;
pq.push(Point(0, 0, A[0][0]));
int res = min(A[0][0], A[R - 1][C - 1]);
while (!pq.empty()) {
Point p = pq.top();
pq.pop();
for (int i = 0; i < 4; ++i) {
int r = p.x + dirs[i][0];
int c = p.y + dirs[i][1];
if (r >= 0 && r < R && c >= 0 && c < C && !visited[r][c]) {
res = min(res, p.val);
if (r == R - 1 && c == C - 1) return res;
visited[r][c] = true;
pq.push(Point(r, c, A[r][c]));
}
}
}
return res;
}
private:
struct Point {
int x, y, val;
Point(int _x, int _y, int _val) : x(_x), y(_y), val(_val) {}
bool operator < (const Point& other) const {
return this->val < other.val;
}
};
int dirs[4][2] = {{-1, 0}, {1, 0}, {0, 1}, {0, -1}};
};

## # 日期

2019 年 9 月 23 日 —— 昨夜睡的早，错过了北京的烟火