【题目链接】
先跑出dfs树。
对于两个操作,先把不连通的情况用并查集搞掉。
另外对于操作1,如果G1-G2是非树边,那么形成了环,这种情况肯定是yes。
然后跑Tarjan,对每个操作形成的子树讨论一下就完了。
#include <cstdio> /* Pigonometry */ #include <cstring> #include <algorithm> #include <vector> using namespace std; const int maxn = 100005, maxm = 500005, maxc = 300005, maxs = 300000; int n, m, head[maxm], cnt, pre[maxn], fa[maxn]; int ans[maxc], A[maxc], B[maxc]; int dfn[maxn], low[maxn], clo, sta[maxs], top; bool cut[maxn], ins[maxn]; struct _edge { int v, next; } g[maxm << 1]; vector<int> c1[maxn], c2[maxn]; inline int iread() { int f = 1, x = 0; char ch = getchar(); for(; ch < '0' || ch > '9'; ch = getchar()) f = ch == '-' ? -1 : 1; for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0'; return f * x; } inline void add(int u, int v) { g[cnt] = (_edge){v, head[u]}; head[u] = cnt++; } inline int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); } inline void dfs(int x) { for(int i = head[x]; ~i; i = g[i].next) if(!pre[g[i].v]) { pre[g[i].v] = x; dfs(g[i].v); } } inline bool check(int x, int u, int v) { if(find(u) == find(v)) return 1; if((!dfn[u] || dfn[u] < dfn[x]) && (!dfn[v] || dfn[v] < dfn[x])) return 1; return 0; } inline void tarjan(int x) { dfn[x] = low[x] = ++clo; ins[sta[++top] = x] = 1; for(int i = head[x]; ~i; i = g[i].next) if(g[i].v ^ pre[x]) { int v = g[i].v; if(!dfn[v]) { tarjan(v); low[x] = min(low[x], low[v]); if(low[v] >= dfn[x]) cut[v] = 1; } else if(ins[v]) low[x] = min(low[x], dfn[v]); } for(int i = 0; i < c2[x].size(); i++) { int u = A[c2[x][i]], v = B[c2[x][i]]; if(check(x, u, v) && u != x && v != x) continue; if(u == x || v == x) ans[c2[x][i]] = 0; if(dfn[u] > dfn[x] && cut[find(u)]) ans[c2[x][i]] = 0; if(dfn[v] > dfn[x] && cut[find(v)]) ans[c2[x][i]] = 0; } for(int i = head[x]; ~i; i = g[i].next) if(dfn[g[i].v] > dfn[x]) fa[find(g[i].v)] = find(x); if(dfn[x] == low[x]) { for(int i = 0; i < c1[x].size(); i++) { int u = A[c1[x][i]], v = B[c1[x][i]]; if(check(x, u, v)) continue; ans[c1[x][i]] = 0; } while(1) { int u = sta[top--]; ins[u] = 0; if(u == x) break; } } } int main() { freopen("2.in", "r", stdin); freopen("2.out", "w", stdout); n = iread(); m = iread(); for(int i = 1; i <= n; i++) head[i] = -1, fa[i] = i; cnt = top = 0; for(int i = 1; i <= m; i++) { int u = iread(), v = iread(); add(u, v); add(v, u); fa[find(u)] = find(v); } for(int i = 1; i <= n; i++) if(!pre[i]) pre[i] = -1, dfs(i); int T = iread(); for(int i = 1; i <= T; i++) { int opt = iread(); A[i] = iread(); B[i] = iread(); ans[i] = 1; if(opt == 1) { int u = iread(), v = iread(); if(find(A[i]) != find(B[i])) { ans[i] = 0; continue; } if(pre[v] != u) swap(v, u); if(pre[v] != u) continue; c1[v].push_back(i); } else { int u = iread(); if(find(A[i]) != find(B[i])) { ans[i] = 0; continue; } c2[u].push_back(i); } } for(int i = 1; i <= n; i++) fa[i] = i; for(int i = 1; i <= n; i++) if(!~pre[i]) tarjan(i); for(int i = 1; i <= T; i++) printf(ans[i] ? "yes\n" : "no\n"); return 0; }