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
cpp
#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：
cpp
#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。
cpp
#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
cpp
#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;
}