标签：int leetcode following matrix height pre volume note cto
Given an m x n matrix of positive integers representing the height of each unit cell in a 2D elevation map, compute the volume of water it is able to trap after raining.
Note:
Both m and n are less than 110. The height of each unit cell is greater than 0 and is less than 20,000.
Example:
Given the following 3x6 height map: [ [1,4,3,1,3,2], [3,2,1,3,2,4], [2,3,3,2,3,1] ] Return 4.
The above image represents the elevation map [[1,4,3,1,3,2],[3,2,1,3,2,4],[2,3,3,2,3,1]] before the rain.
After the rain, water are trapped between the blocks. The total volume of water trapped is 4.
解题思路：这题和"关门打狗"的想法是一样的，首先把最外面一圈围起来,利用一个优先队列每次从最矮的高度出发，往里面缩小范围，如果碰到更矮的，则此处可以积水当前高度-更矮的高度，如果碰到比自己高的，则把自己的高度变成高的高度，积水体积不变。
class Solution { public: int trapRainWater(vector<vector<int>>& heightMap) { if(heightMap.empty())return 0; priority_queue<pair<int,int>, vector<pair<int,int> >, greater<pair<int,int> > >q; int n=heightMap.size(),m=heightMap[0].size(); vector<vector<int> >vis(n,vector<int>(m,0)); for(int i=0;i<n;i++){ vis[i][0]=vis[i][m-1]=1; q.push(make_pair(heightMap[i][0],i*m)); q.push(make_pair(heightMap[i][m-1],i*m+m-1)); } for(int i=0;i<m;i++){ vis[0][i]=vis[n-1][i]=1; q.push(make_pair(heightMap[0][i],i)); q.push(make_pair(heightMap[n-1][i],(n-1)*m+i)); } vector<vector<int> >dirs{{0,1},{1,0},{0,-1},{-1,0}}; int res=0; while(!q.empty()){ auto val=q.top(); q.pop(); int height=val.first,x=val.second/m,y=val.second%m; for(auto dir: dirs){ int xx=x+dir[0],yy=y+dir[1]; if(xx<0||xx>=n||yy<0||yy>=m||vis[xx][yy])continue; vis[xx][yy]=1; res += max(0,height-heightMap[xx][yy]); q.push(make_pair(heightMap[xx][yy],xx*m+yy)); } } return res; } };
标签：int leetcode following matrix height pre volume note cto
原文地址：http://www.cnblogs.com/tsunami-lj/p/6527199.html
