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算法训练营第六十七天 | 卡码网110 字符串接龙、卡码网105 有向图的完全可达性、卡码网106 岛屿的周长

news 来源：原创 2024/6/30 23:07:23

卡码网110 字符串接龙

这题一开始用的邻接表+dfs，不幸超时

#include <iostream>
#include <list>
#include <string>
#include <vector>
using namespace std;int minLen = 501;bool count(string a, string b) {int num = 0;for (int i = 0; i < a.length() && i < b. length(); i++) {if (a[i] != b[i]) {num++;if (num > 1) return false;}}return true;
}int dfs(string* s, vector<list<int>>& l, int* visited, int len, int index, int n) {if (index == n + 1) {if (minLen > len) minLen = len;return len;}int result = 501;for (auto it = l[index].begin(); it != l[index].end(); it++) {if (!visited[*it]) {visited[*it] = 1;result = dfs(s, l, visited, len+1, *it, n);visited[*it] = 0;}}return min(minLen, result);
}int main() {int n;cin >> n;string s[n+2];cin >> s[0] >> s[n+1];for (int i = 1; i <= n; i++) {cin >> s[i];}vector<list<int>> l(n+1);for (int i = 0; i <= n; i++) {for (int j = 1; j <= n+1; j++) {if (j != i) {if (count(s[i], s[j])) l[i].push_back(j);}}}int visited[n+2];for (int i = 0; i < n+2; i++) visited[i] = 0;visited[0] = 1;minLen =  dfs(s, l, visited, 1, 0, n);if (minLen == 501) cout << 0 << endl;else cout << minLen << endl;
}

题解中用的广搜，可以去看下，说是最短路径用深搜比较麻烦，要确定哪条最短，我确定出来了，但还是超时了。广搜只要搜到就一定是最短路径了，先放这。

卡码网105 有向图的完全可达性

这题就比较轻松了，深搜也不需要完全回溯，只需要遍历记录即可

代码如下：

#include <iostream>
#include <list>
#include <vector>
#include <queue>
using namespace std;void bfs(vector<list<int>>& l, vector<int>& visited, int index) {queue<int> myque;myque.push(index);while (!myque.empty()) {for (int i : l[index]) {if (!visited[i]) {visited[i] = 1;myque.push(i);}}myque.pop();}
}int main() {int n, k;cin >> n >> k;int s, t;vector<list<int>> l(n);for (int i = 0; i < k; i++) {cin >> s >> t;l[s-1].push_back(t-1);}vector<int> visited(n, 0);visited[0] = 1;bfs(l, visited, 0);for (int i = 0; i < n; i++) {if (!visited[i]) {cout << -1 << endl;return 0;}}cout << 1 << endl;return 0;
}

卡码网106 岛屿的周长

这题很简单，没用到dfs和bfs，简单填充边界后判断周围是否触碰边界后计数即可。

代码如下：

#include <iostream>using namespace std;int main() {int n, m;cin >> n >> m;int g[n][m];int sum = 0;for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {cin >> g[n][m];}}for (int i = 1; i < n - 1; i++) {for (int j = 1; j < m - 1; j++) {if (g[i][j] == 1) {if (g[i-1][j] == 0) sum++;if (g[i][j-1] == 0) sum++;if (g[i+1][j] == 0) sum++;if (g[i][j+1] == 0) sum++;}}}cout << sum << endl;
}