在这个物欲横流的社会
oj冷漠无情
只有这xx还有些温度
越界就越界吧 wrong 怎么回事。。。。
给出一个图
然后给出q次询问
问是否存在v和w分别到u的路径且边不重复
在边双连通分量中 任意两点 都至少存在两条边不重复的路径
那么若u v w 在同一连通图中
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则存在
若不在
则只有u在 v 和 w 中间时才存在
想一想是不是
#include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <map> #include <cctype> #include <set> #include <vector> #include <stack> #include <queue> #include <algorithm> #include <cmath> #include <bitset> #define rap(i, a, n) for(int i=a; i<=n; i++) #define rep(i, a, n) for(int i=a; i<n; i++) #define lap(i, a, n) for(int i=n; i>=a; i--) #define lep(i, a, n) for(int i=n; i>a; i--) #define rd(a) scanf("%d", &a) #define rlld(a) scanf("%lld", &a) #define rc(a) scanf("%c", &a) #define rs(a) scanf("%s", a) #define rb(a) scanf("%lf", &a) #define rf(a) scanf("%f", &a) #define pd(a) printf("%d\n", a) #define plld(a) printf("%lld\n", a) #define pc(a) printf("%c\n", a) #define ps(a) printf("%s\n", a) #define MOD 2018 #define LL long long #define ULL unsigned long long #define Pair pair<int, int> #define mem(a, b) memset(a, b, sizeof(a)) #define _ ios_base::sync_with_stdio(0),cin.tie(0) //freopen("1.txt", "r", stdin); using namespace std; const int maxn = 101000, INF = 0x7fffffff, maxm = 1000100, maxh = 20; int n, m, q; int head[maxn], nex[maxm], cnt, head2[maxn], cnt2; int pre[maxn], low[maxn], sccno[maxn], dfs_clock, scc_cnt; int pa[maxn], anc[maxn][maxh + 1], deep[maxn], instack[maxn]; int f[maxn]; stack<int> s; struct node { int u, v; }Node[maxm]; void add_(int u, int v) { Node[cnt].u = u; Node[cnt].v = v; nex[cnt] = head[u]; head[u] = cnt++; } void add(int u, int v) { add_(u, v); add_(v, u); } struct edge{ int u, v, next; }Edge[maxm]; void add2(int u, int v) { Edge[cnt2].u = u; Edge[cnt2].v = v; Edge[cnt2].next = head2[u]; head2[u] = cnt2++; } void tarjan(int u,int fa){ low[u] = pre[u] = ++dfs_clock; s.push(u); instack[u] = 1; for(int i = head[u]; i != -1; i = nex[i]) { int v = Node[i].v; if(i == (fa ^ 1)) continue; if(!pre[v]) { tarjan(v, i); low[u] = min(low[u], low[v]); } else if(instack[v]) low[u] = min(low[u], pre[v]); } if(pre[u] == low[u]) { scc_cnt++; for(;;) { int x = s.top(); s.pop(); sccno[x] = scc_cnt; instack[x] = 0; if(x == u) break; } } } int dfs(int u, int fa) { for(int i = 1; i < maxh; i++) anc[u][i] = anc[anc[u][i - 1]][i - 1]; for(int i = head2[u]; i != -1; i = Edge[i].next) { int v = Edge[i].v; if(v == fa || deep[v]) continue; anc[v][0] = u; deep[v] = deep[u] + 1; dfs(v, u); } } int lca(int u, int v) { if(deep[u] < deep[v]) swap(u, v); for(int i = maxh; i >= 0; i--) if(deep[anc[u][i]] >= deep[v]) u = anc[u][i]; for(int i = maxh; i >= 0; i--) { if(anc[u][i] != anc[v][i]) { u = anc[u][i]; v = anc[v][i]; } } if(u == v) return u; return anc[u][0]; } int find(int x) { return f[x] == x ? x : (f[x] = find(f[x])); } void init() { rep(i, 1, maxn) f[i] = i; mem(head, -1); mem(head2, -1); mem(deep, 0); dfs_clock = scc_cnt = cnt = cnt2 = 0; mem(sccno, 0); mem(pre, 0); mem(low, 0); mem(anc, 0); mem(instack, 0); } int main() { int T; rd(T); while(T--) { init(); rd(n), rd(m), rd(q); rap(i, 1, m) { int u, v; rd(u), rd(v); if(u == v) continue; add(u, v); } for(int i = 1; i <= n; i++) if(!pre[i]) tarjan(i, -1); rap(u, 1, n) { for(int i = head[u]; i != -1; i = nex[i]) { int v = Node[i].v; if(sccno[u] != sccno[v]) { add2(sccno[u], sccno[v]); int l = find(sccno[u]), r = find(sccno[v]); if(l == r) continue; f[l] = r; } } } rap(i, 1, scc_cnt) if(!deep[i]) deep[i] = 1, dfs(i, -1); rap(i, 1, q) { int u, v, w; rd(u), rd(v), rd(w); if(find(sccno[u]) != find(sccno[v]) || find(sccno[u]) != find(sccno[w])) { puts("No"); continue; } if(sccno[v] != sccno[w]) { int lc = lca(sccno[v], sccno[w]); int ans1 = lca(sccno[u], sccno[v]); int ans2 = lca(sccno[u], sccno[w]); int ans3 =lca(lc, sccno[u]); if(ans3 == lc && (ans1 == sccno[u] || ans2 == sccno[u])) { puts("Yes"); } else { puts("No"); } } else { if(sccno[u] == sccno[v] ) { puts("Yes"); continue; } else puts("No"); } } } return 0; }