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Given n x m non-negative integers representing an elevation map 2d where the area of each cell is 1 x 1, compute how much water it is able to trap after raining.
Given 5*4 matrix
[12,13,0,12]
[13,4,13,12]
[13,8,10,12]
[12,13,12,12]
[13,13,13,13]
return 14.
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最小化堆。注意在往堆里放新元素的时候，要同时更新位置的水位高度。即：
heap.emplace(xx, yy, max(u.h, heights[xx][yy]));
还要注意优先队列默认是大顶堆，默认使用<号，所以要想使用小顶堆，在自己写cmp仿函数时，要返回a > b, 还不是a < b。
1 class Solution { 2 public: 3 /** 4 * @param heights: a matrix of integers 5 * @return: an integer 6 */ 7 struct cell { 8 int x, y; 9 int h; 10 cell() {} 11 cell(int _x, int _y, int _h) : x(_x), y(_y), h(_h) {} 12 }; 13 struct cmp { 14 bool operator() (const cell &a, const cell &b) { 15 return a.h > b.h; 16 } 17 }; 18 int trapRainWater(vector<vector<int> > &heights) { 19 // write your code here 20 if (heights.empty() || heights[0].empty()) return 0; 21 int n = heights.size(), m = heights[0].size(); 22 vector<vector<bool>> visit(n, vector<bool>(m, false)); 23 priority_queue<cell, vector<cell>, cmp> heap; 24 for (int i = 0; i < n; ++i) { 25 heap.emplace(i, 0, heights[i][0]); 26 heap.emplace(i, m - 1, heights[i][m-1]); 27 visit[i][0] = visit[i][m-1] = true; 28 } 29 for (int j = 0; j < m; ++j) { 30 heap.emplace(0, j, heights[0][j]); 31 heap.emplace(n - 1, j, heights[n-1][j]); 32 visit[0][j] = visit[n-1][j] = true; 33 } 34 int res = 0; 35 const int dx[4] = {0, 1, 0, -1}; 36 const int dy[4] = {1, 0, -1, 0}; 37 while (!heap.empty()) { 38 auto u = heap.top(); 39 heap.pop(); 40 for (int i = 0; i < 4; ++i) { 41 int xx = u.x + dx[i], yy = u.y + dy[i]; 42 if (xx >= 0 && xx < n && yy >= 0 && yy < m && !visit[xx][yy]) { 43 res += max(0, u.h - heights[xx][yy]); 44 heap.emplace(xx, yy, max(u.h, heights[xx][yy])); 45 visit[xx][yy] = true; 46 } 47 } 48 } 49 return res; 50 } 51 };
[LintCode] Trapping Rain Water II
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原文地址：http://www.cnblogs.com/easonliu/p/4743644.html
