101. 孤岛的总面积
先用dfs遍历四条边界，把与之相连的置为0
再用计算孤岛面积的方法统计总面积
cpp
#include <iostream>
#include <vector>
using namespace std;
int dir[4][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
void dfs(vector<vector<int>>& graph, int x, int y){
graph[x][y] = 0;
for(int i = 0; i < 4; i++){
int nextx = x + dir[i][0];
int nexty = y + dir[i][1];
if(nextx < 0 || nextx >= graph.size() || nexty < 0 || nexty >= graph[0].size()){
continue;
}
if(graph[nextx][nexty] == 1){
dfs(graph, nextx, nexty);
}
}
}
void dfs(vector<vector<int>>& graph, int x, int y, vector<vector<bool>>& visited, int &res){
if(visited[x][y] == true){
return;
}
res++;
visited[x][y] = true;
for(int i = 0; i < 4; i++){
int nextx = x + dir[i][0];
int nexty = y + dir[i][1];
if(nextx < 0 || nextx >= graph.size() || nexty < 0 || nexty >= graph[0].size()){
continue;
}
if(graph[nextx][nexty] == 1){
dfs(graph, nextx, nexty, visited, res);
}
}
}
int main(){
int N, 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);
}
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);
}
if(graph[N - 1][j] == 1){
dfs(graph, N - 1, j);
}
}
int res = 0;
vector<vector<bool>> visited(N, vector<bool>(M, false));
for(int i = 0; i < N; i++){
for(int j = 0; j < M; j++){
if(graph[i][j] == 1){
dfs(graph, i, j, visited, res);
}
}
}
cout << res;
return 0;
}102. 沉没孤岛
先将四条边相连的地块都置为2，最后再全部遍历，将为1 为0的都置零，为2的置为1
cpp
#include <iostream>
#include <vector>
using namespace std;
int dir[4][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
void dfs(vector<vector<int>>& graph, int x, int y){
graph[x][y] = 2;
for(int i = 0; i < 4; i++){
int nextx = x + dir[i][0];
int nexty = y + dir[i][1];
if(nextx < 0 || nextx >= graph.size() || nexty < 0 || nexty >= graph[0].size()){
continue;
}
if(graph[nextx][nexty] == 1){
dfs(graph, nextx, nexty);
}
}
}
int main(){
int N, 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);
}
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);
}
if(graph[N - 1][j] == 1){
dfs(graph, N - 1, j);
}
}
for(int i = 0; i < N; i++){
for(int j = 0; j < M - 1; j++){
if(graph[i][j] == 1){
graph[i][j] = 0;
}
if(graph[i][j] == 2){
graph[i][j] = 1;
}
cout << graph[i][j] << " ";
}
if(graph[i][M - 1] == 2){
graph[i][M - 1] = 1;
}
cout << graph[i][M - 1] << endl;
}
return 0;
}103. 高山流水
可以从边往中间搜，这样只用搜索四条边，不需要全部遍历
cpp
#include<iostream>
#include<vector>
using namespace std;
int dir[4][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
void dfs(vector<vector<int>>& graph, vector<vector<bool>>& Border, int x, int y){
if(Border[x][y]){
return;
}
Border[x][y] = true;
for(int i = 0; i < 4; i++){
int nextx = x + dir[i][0];
int nexty = y + dir[i][1];
if(nextx < 0 || nextx >= graph.size() || nexty < 0 || nexty >= graph[0].size()){
continue;
}
if(graph[nextx][nexty] >= graph[x][y]){
dfs(graph, Border, nextx, nexty);
}
}
}
int main(){
int N, 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];
}
}
vector<vector<bool>> firstBorder(N, vector<bool>(M, false));
vector<vector<bool>> secondBorder(N, vector<bool>(M, false));
for(int i = 0; i < N; i++){
dfs(graph, firstBorder, i, 0);
dfs(graph, secondBorder, i, M - 1);
}
for(int j = 0; j < M; j++){
dfs(graph, firstBorder, 0, j);
dfs(graph, secondBorder, N - 1, j);
}
for(int i = 0; i < N; i++){
for(int j = 0; j < M; j++){
if(firstBorder[i][j] && secondBorder[i][j]){
cout << i << " " << j << endl;
}
}
}
return 0;
}104. 建造最大岛屿
遍历二维网格，通过将一个海洋格子（0）改为陆地（1），找到改造后能形成的最大岛屿面积（岛屿为上下左右相邻的陆地）。
标记岛屿 & 统计面积
遍历网格，对每个未标记的陆地（1），用 DFS 遍历其连通区域，将该区域所有格子标记为唯一岛屿编号（从 2 开始）；
记录每个岛屿编号对应的面积，同时保存原始最大岛屿面积（无改造时的最大值）。
计算改造海洋的最大收益
遍历每个海洋格子（0），统计其四个方向相邻的不同岛屿面积之和（用集合去重，避免重复加同一岛屿）；
该和 + 1（改造的海洋格子本身）即为改造后岛屿面积，更新全局最大面积。
初始最大面积设为 1（应对全海洋场景：改造 1 个格子成陆地，面积为 1）；
遍历海洋时先做边界检查，避免数组越界。
cpp
#include <vector>
#include <iostream>
#include <unordered_map>
#include <unordered_set>
#include <algorithm>
using namespace std;
int dir[4][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
void dfs(vector<vector<int>>& graph, int x, int y, int &areaSum, int islandId){
graph[x][y] = islandId;
areaSum++;
for(int i = 0; i < 4; i++){
int nextx = x + dir[i][0];
int nexty = y + dir[i][1];
if(nextx < 0 || nexty < 0 || nextx >= graph.size() || nexty >= graph[0].size()){
continue;
}
if(graph[nextx][nexty] == 1){
dfs(graph, nextx, nexty, areaSum, islandId);
}
}
}
int main(){
int N, 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];
}
}
int islandId = 2;
int res = 1;
unordered_map<int, int> mp;
for(int i = 0; i < N; i++){
for(int j = 0; j < M; j++){
if(graph[i][j] == 1){
int sum = 0;
dfs(graph, i, j, sum, islandId);
mp[islandId] = sum;
islandId++;
res = max(res, sum);
}
}
}
for(int i = 0; i < N; i++){
for(int j = 0; j < M; j++){
if(graph[i][j] == 0){
int curSum = 1;
unordered_set<int> st;
for(int k = 0; k < 4; k++){
int nextx = i + dir[k][0];
int nexty = j + dir[k][1];
if(nextx < 0 || nexty < 0 || nextx >= graph.size() || nexty >= graph[0].size()){
continue;
}
if(graph[nextx][nexty] == 0 || st.count(graph[nextx][nexty]) != 0) {
continue;
}
else {
curSum += mp[graph[nextx][nexty]];
res = max(res, curSum);
st.insert(graph[nextx][nexty]);
}
}
}
}
}
cout << res;
return 0;
}