接雨水 II-LintCode

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.

例如，给定一个 5*4 的矩阵：

[
[12,13,0,12],
[13,4,13,12],
[13,8,10,12],
[12,13,12,12],
[13,13,13,13]
]
返回 14.

#ifndef C364_H#define C364_H#include<iostream>#include<vector>#include<functional>#include<queue>using namespace std;class Solution {public:    /*     * @param heights: a matrix of integers     * @return: an integer     */    int trapRainWater(vector<vector<int>> &heights) {        // write your code here        if (heights.empty())            return 0;        int num = 0;         int Max = 0;        priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> que;        int row = heights.size();        int col = heights[0].size();        vector<vector<int>> visited(row, vector<int>(col, 0));        for (int i = 0; i < row; ++i)        {            for (int j = 0; j < col; ++j)            {                if (!(i == 0 || j == 0 || i == row - 1 || j == col - 1))                    continue;                visited[i][j] = 1;                que.push(make_pair(heights[i][j], i*col + j));            }        }        while (!que.empty())        {            auto pairVal = que.top();            que.pop();            int height = pairVal.first;            Max = max(height, Max);            int xp = pairVal.second / col;            int yp = pairVal.second%col;            vector<vector<int>> boundry{ { -1, 0 }, { 0, -1 }, { 1, 0 }, { 0, 1 } };            for (auto c : boundry)            {                int Xp = xp + c[0];                int Yp = yp + c[1];                if (Xp < 0 || Yp < 0 || Xp >= row || Yp >= col||visited[Xp][Yp])                    continue;                visited[Xp][Yp] = 1;                que.push(make_pair(heights[Xp][Yp], Xp*col + Yp));                if (heights[Xp][Yp] < Max)                    num += Max - heights[Xp][Yp];            }        }        return num;    }};#endif