代码随想录算法训练营第58天 [101.孤岛的总面积 102.沉没孤岛 103.水流问题 104.建造最大岛屿]
一、101.孤岛的总面积
链接: 代码随想录.
思路:从四个边缘开始深搜,搜到了标记为2,这样所有为2和0的就不是孤岛,所有为1的就是孤岛
做题状态:看解析后做出来了
#include<iostream>
#include<vector>
using namespace std;
int dir[4][2] = {{0,1},{0,-1},{1,0},{-1,0}};void dfs(vector<vector<int>> &graph,int x,int y){graph[x][y] = 0;for(int i = 0;i<4;i++){int next_x = x+dir[i][0];int next_y = y+dir[i][1];if(next_x<0 || next_x>=graph.size()||next_y<0||next_y>=graph[0].size()){continue;}if(graph[next_x][next_y] == 1){dfs(graph,next_x,next_y);}}
}int main(){int n;int m;cin >> n >> m;vector<vector<int>> graph(n,vector<int>(m,0));for(int i = 0;i<n;i++){for(int j = 0;j<m;j++){cin >> graph[i][j];}}for(int i = 0;i<n;i++){if(graph[i][0] == 1)dfs(graph,i,0);}for(int i = 0;i<n;i++){if(graph[i][m-1] == 1)dfs(graph,i,m-1);}for(int j = 0;j<m;j++){if(graph[0][j] == 1)dfs(graph,0,j);}for(int j = 0;j<m;j++){if(graph[n-1][j] == 1)dfs(graph,n-1,j);}int result = 0;for(int i = 0;i<n;i++){for(int j = 0;j<m;j++){// cout << graph[i][j] << " ";if(graph[i][j] == 1){result++;} }// cout << endl;}cout << result <<endl;
}
二、102.沉没孤岛
链接: 代码随想录.
思路:从四个边缘开始深搜,搜到了标记为2,这样所有为2和0的就不是孤岛,所有为1的就是孤岛,最后把2变为1,把1变为0
做题状态:看解析后做出来了
#include<iostream>
#include<vector>
using namespace std;
int dir[4][2] = {{0,1},{0,-1},{1,0},{-1,0}};void dfs(vector<vector<int>> &graph,int x,int y){graph[x][y] = 2;for(int i = 0;i<4;i++){int next_x = x+dir[i][0];int next_y = y+dir[i][1];if(next_x<0 || next_x>=graph.size()||next_y<0||next_y>=graph[0].size()){continue;}if(graph[next_x][next_y] == 1){dfs(graph,next_x,next_y);}}
}int main(){int n;int m;cin >> n >> m;vector<vector<int>> graph(n,vector<int>(m,0));for(int i = 0;i<n;i++){for(int j = 0;j<m;j++){cin >> graph[i][j];}}for(int i = 0;i<n;i++){if(graph[i][0] == 1)dfs(graph,i,0);}for(int i = 0;i<n;i++){if(graph[i][m-1] == 1)dfs(graph,i,m-1);}for(int j = 0;j<m;j++){if(graph[0][j] == 1)dfs(graph,0,j);}for(int j = 0;j<m;j++){if(graph[n-1][j] == 1)dfs(graph,n-1,j);}for(int i = 0;i<n;i++){for(int j = 0;j<m;j++){if(graph[i][j] == 1){graph[i][j] = 0;} if(graph[i][j] == 2){graph[i][j] = 1;} cout << graph[i][j] << " ";}cout << endl;}
}
三、103.水流问题
链接: 代码随想录.
思路: 某个点能不能流到第一组边界和第二组边界 转化为 第一组边界和第二组边界能不能流到某个点
做题状态:看解析后做出来了
#include<iostream>
#include<vector>
using namespace std;
int dir[4][2] = {{0,1},{0,-1},{1,0},{-1,0}};void dfs(vector<vector<int>> &graph,vector<vector<int>> &visited,int x,int y){if(visited[x][y]){return;}visited[x][y] = 1;for(int i = 0;i<4;i++){int next_x = x+dir[i][0];int next_y = y+dir[i][1];if(next_x<0 || next_x>=graph.size()||next_y<0||next_y>=graph[0].size()){continue;}if(graph[x][y] <= graph[next_x][next_y]){dfs(graph,visited,next_x,next_y);}}
}int main(){int n;int m;cin >> n >> m;vector<vector<int>> graph(n,vector<int>(m,0));for(int i = 0;i<n;i++){for(int j = 0;j<m;j++){cin >> graph[i][j];}}//某个点能不能流到第一组边界和第二组边界//转化为//第一组边界和第二组边界能不能流到某个点vector<vector<int>> first_visited(n,vector<int>(m,0));vector<vector<int>> second_visited(n,vector<int>(m,0));for(int i = 0;i<n;i++){dfs(graph,first_visited,i,0);dfs(graph,second_visited,i,m-1);}for(int j = 0;j<m;j++){dfs(graph,first_visited,0,j);dfs(graph,second_visited,n-1,j);}vector<vector<int>> result(n,vector<int>(m,0));for(int i = 0;i<n;i++){for(int j = 0;j<m;j++){// cout << second_visited[i][j] <<" ";if(first_visited[i][j] == 1 && second_visited[i][j] == 1){cout << i << " " << j <<endl;}}// cout << endl;}}
四、104.建造最大岛屿
链接: 代码随想录.
思路:给不同的岛屿标记不同的编号mark,再遍历所有为海水的区域将其变为陆地,计算联通起来的最大面积
做题状态:看解析后做出来了
#include <iostream>
#include <vector>
#include <unordered_set>
#include <unordered_map>
using namespace std;
int n;
int m;
int count = 0;int dir[4][2] = {{0,1},{0,-1},{1,0},{-1,0}};void dfs(vector<vector<int>> &graph,vector<vector<int>> &visited,int x,int y,int mark){if(visited[x][y] == 1 || graph[x][y] == 0){return;}visited[x][y] = 1;graph[x][y] = mark;count++;for(int i = 0;i<4;i++){int next_x = x+dir[i][0];int next_y = y+dir[i][1];if(next_x<0 || next_x>=n ||next_y< 0|| next_y>=m){continue;}dfs(graph,visited,next_x,next_y,mark);}
}int main(){cin >> n >>m;vector<vector<int>> graph(n,vector<int>(m,0));vector<vector<int>> visited(n,vector<int>(m,0));for(int i = 0;i<n;i++){for(int j = 0;j<m;j++){cin >> graph[i][j];}}unordered_map<int,int> gridNum;int mark = 2;bool isAllGrid = true;for(int i = 0;i<n;i++){for(int j = 0;j<m;j++){if(graph[i][j] == 0){isAllGrid = false;}if(graph[i][j] == 1 && visited[i][j] == 0){count = 0;dfs(graph,visited,i,j,mark);gridNum[mark] = count;mark++;}}}if(isAllGrid){cout <<m*n<<endl;return 0;}int result = 0;unordered_set<int> visitedGraph;for(int i = 0;i<n;i++){for(int j = 0;j<m;j++){count = 1;visitedGraph.clear();if(graph[i][j] == 0){for(int k = 0;k<4;k++){int next_x = i+dir[k][0];int next_y = j+dir[k][1]; if(next_x<0 || next_x>=n ||next_y< 0|| next_y>=m){continue;}if(visitedGraph.count(graph[next_x][next_y])){continue;}count += gridNum[graph[next_x][next_y]];visitedGraph.insert(graph[next_x][next_y]);}}result = max(result,count);}}cout << result <<endl;}