lanqiao3517 砍树(DFS)

14lanqiao/test10-1.cpp

@@ -0,0 +1,62 @@
+// lanqiao3517 砍树(DFS)
+#include<bits/stdc++.h>
+using namespace std;
+#define int long long
+const int N = 1e5 + 9;
+vector<pair<int, int>> g[N];
+bool vis[N];
+int n, m, s[N], t[N];
+// 暴力解法: 枚举删除的边, 利用dfs查询数据对之间的连通性-时间复杂组为O(n^2m) 
+bool dfs(int s, int t, int id){
+    if(s == t){
+        return true;
+    }
+    for(int i = 0; i < g[s].size(); i++){
+        int nxt = g[s][i].first, edgeId = g[s][i].second;
+        if(edgeId == id) continue;
+        if(vis[nxt]) continue;
+        vis[nxt] = 1;
+        if(dfs(nxt, t, id)) return true;
+    }
+    return false;
+}
+signed main(){
+    cin >> n >> m;
+    for(int i = 1; i < n; i++){
+        int u, v; cin >> u >> v;
+        g[u].push_back({v, i});
+        g[v].push_back({u, i});
+    }
+    for(int i = 1; i <= m; i++){
+        cin >> s[i] >> t[i];
+    }
+    // 枚举需要删除的边的编号-从大到小枚举
+    for(int i = n - 1; i >= 1; i--){
+        bool flag = true;
+        for(int j = 1; j <= m; j++){
+            memset(vis, 0, sizeof vis);
+            vis[s[j]] = 1;
+            if(dfs(s[j], t[j], i)){
+                flag = false;
+                break;
+            }
+        }
+        if(flag){
+            cout << i << ' ';
+            return 0;
+        }
+    }
+
+    cout << -1 << endl;
+    return 0;
+}
+/* test samples -> 4
+6 2
+1 2
+2 3
+4 3
+2 5
+6 5
+3 6
+4 5
+*/