110.字符串接龙
题目链接：110. 字符串接龙文章讲解：代码随想录
思路：
把每个字符串看成图的一个节点。
转换为求无权图两节点的的最短路径。求最短路径用bfs
#include <string>
#include <vector>
#include <iostream>
#include <queue>
#include <unordered_map>
using namespace std;
unordered_map<int, int>mymap;bool canTransform(string a, string b) {int count = 0;if (a.size() != b.size())return false;int size = min(a.size(), b.size());for (int i = 0; i < size; i++) {if (a[i] != b[i])count++;}if (count == 1)return true;else return false;
}//广搜求最短路径
int bfs(vector<vector<bool>>graph, int begin, int end) {queue<int>mq;mq.push(begin);while (!mq.empty()) {int curStr = mq.front();if (curStr == end) { return mymap[end]; }mq.pop();for (int i = 0; i < graph.size(); i++) {if (graph[curStr][i] == true && !mymap.count(i)) {//mymap.count(i)防止走回头路mymap[i] = mymap[curStr] + 1;mq.push(i);}}}return 0;
}int main() {//数据读取mymap[0] = 1; //初始化不能在全局领域初始化int n;string beginStr, endStr;cin >> n;cin >> beginStr >> endStr;vector<string>strList;strList.push_back(beginStr);int size = n; //使用while（n--）会改变n的值while (size--) {string str;cin >> str;strList.push_back(str);}strList.push_back(endStr);//构造图vector<vector<bool>>graph(n + 2, vector<bool>(n + 2, false));for (int i = 0; i < graph.size(); i++) {for (int j = 0; j < graph.size(); j++) {if (canTransform(strList[i], strList[j]))graph[i][j] = true;}}int ans = 0;ans = bfs(graph, 0, n + 1);cout << ans;
}105.有向图的完全可达性
题目链接：105. 有向图的完全联通
文章讲解：代码随想录
思路：
用深搜
逐渐遍历看第一个节点能不能到达其他节点
错误深搜：
#include <iostream>
#include <vector>
using namespace std;bool dfs(vector<pair<int, int>>graph, int begin, int end) {if (begin == end)return true;for (int i = 0; i < graph.size(); i++) {int first = graph[i].first;int second = graph[i].second;if (first == begin) {if (dfs(graph, second, end))break;}}return false;
}int main() {int n, k;cin >> n >> k;vector<pair<int, int>>graph;for (int i = 0; i < k; i++) {int s, t;cin >> s >> t;graph.push_back({ s,t });}int ans = 1;for (int i = 2; i <= n; i++) {if (!dfs(graph, 1, i)) { ans = -1; }}cout << ans;}错误原因：
没有visited记录已经访问过的节点 ，导致陷入死循环。
#include <iostream>
#include <vector>
using namespace std;bool dfs(vector<pair<int, int>>graph,vector<bool>&visited, int begin, int end) {visited[begin]=true; //visited记录已经访问过的节点if (begin == end)return true;for (int i = 0; i < graph.size(); i++) {int first = graph[i].first;int second = graph[i].second;if (first == begin&&visited[second]==false) {if(dfs(graph,visited, second, end))return true;}}return false;
}int main() {int n, k;cin >> n >> k;vector<pair<int, int>>graph;for (int i = 0; i < k; i++) {int s, t;cin >> s >> t;graph.push_back({ s,t });}int ans = 1;for (int i = 2; i <= n; i++) {vector<bool>visited(n,false);if (!dfs(graph, visited,1, i)) { ans = -1; }}cout << ans;}106.岛屿的周长
题目链接：106. 岛屿的周长
文章讲解：代码随想录
思路：
遍历所有陆地，统计其四个方向 如果是海 则边数+1
不要用惯性思维用深搜
#include <iostream>
#include <vector>
using namespace std;
int dir[4][2]={{-1,0},{1,0},{0,1},{0,-1}};
int main(){ int n,m;cin>>n>>m;vector<vector<int>>grid(n,vector<int>(m,0));for(int i=0;i<n;i++){for(int j=0;j<m;j++){cin>>grid[i][j];}}int ans=0;for(int i=0;i<n;i++){for(int j=0;j<m;j++){if(grid[i][j]==1){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>=grid.size()||nexty>=grid[0].size()){ans+=1;continue;}if(grid[nextx][nexty]==0)ans++;}}}}cout<<ans;
}
)
