写这道题写的真是被精神污染了。。。主席树什么的全都不会了唔
还是Orz PoPoQQQ吧，原来打算写题解的，写了半个小时发现语文不好2333
1 /************************************************************** 2 Problem: 3772 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:3960 ms 7 Memory:62540 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <algorithm> 12 #include <vector> 13 14 using namespace std; 15 typedef long long ll; 16 const int N = 100010; 17 const int M = N; 18 const int cnt_segment = 3804000; 19 20 struct edge { 21 int next, to; 22 edge() {} 23 edge(int _n, int _t) : next(_n), to(_t) {} 24 } e[N << 1]; 25 26 struct Query { 27 int x, y; 28 Query() {} 29 Query(int _x, int _y) : x(_x), y(_y) {} 30 31 inline bool operator < (const Query &a) const { 32 return x == a.x ? y < a.y : x < a.x; 33 } 34 inline bool operator == (const Query &a) const { 35 return x == a.x && y == a.y; 36 } 37 } Q[M]; 38 39 struct seg_node { 40 seg_node *ls, *rs; 41 int val; 42 } *seg[N], mempool[cnt_segment], *cnt_seg = mempool; 43 44 struct tree_node { 45 int fa[18], dep, st, ed; 46 } tr[N]; 47 48 int n, m; 49 int first[N], tot; 50 int cnt_dfs; 51 ll ans1, ans2, g; 52 vector <int> a[N]; 53 54 inline int read() { 55 int x = 0; 56 char ch = getchar(); 57 while (ch < '0' || '9' < ch) 58 ch = getchar(); 59 while ('0' <= ch && ch <= '9') 60 (x *= 10) += ch - '0', ch = getchar(); 61 return x; 62 } 63 64 inline ll gcd(ll a, ll b) { 65 return a ? gcd(b % a, a) : b; 66 } 67 68 inline void Add_edges(int x, int y) { 69 e[++tot] = edge(first[x], y), first[x] = tot; 70 e[++tot] = edge(first[y], x), first[y] = tot; 71 } 72 73 #define mid (l + r >> 1) 74 void seg_modify(seg_node *&p, int l, int r, int pos, int v) { 75 *(++cnt_seg) = *p, p = cnt_seg, p -> val += v; 76 if (l == r) return; 77 if (pos <= mid) seg_modify(p -> ls, l, mid, pos, v); 78 else seg_modify(p -> rs, mid + 1, r, pos, v); 79 } 80 81 int get_ans(seg_node *p1, seg_node *p2, seg_node *p3, seg_node *p4, int l, int r, int x, int y) { 82 if (l == x && y == r) 83 return p1 -> val + p2 -> val - p3 -> val - p4 -> val; 84 if (y <= mid) 85 return get_ans(p1 -> ls, p2 -> ls, p3 -> ls, p4 -> ls, l, mid, x, y); 86 if (mid < x) 87 return get_ans(p1 -> rs, p2 -> rs, p3 -> rs, p4 -> rs, mid + 1, r, x, y); 88 return get_ans(p1 -> ls, p2 -> ls, p3 -> ls, p4 -> ls, l, mid, x, mid) + 89 get_ans(p1 -> rs, p2 -> rs, p3 -> rs, p4 -> rs, mid + 1, r, mid + 1, y); 90 } 91 #undef mid 92 93 #define y e[x].to 94 void dfs1(int p) { 95 int x; 96 tr[p].st = ++cnt_dfs; 97 for (x = 1; x <= 17; ++x) 98 tr[p].fa[x] = tr[tr[p].fa[x - 1]].fa[x - 1]; 99 for (x = first[p]; x; x = e[x].next) 100 if (y != tr[p].fa[0]) { 101 tr[y].fa[0] = p, tr[y].dep = tr[p].dep + 1; 102 dfs1(y); 103 } 104 tr[p].ed = ++cnt_dfs; 105 } 106 107 void dfs2(int p) { 108 int x; 109 vector <int> :: iterator IT; 110 seg[p] = seg[tr[p].fa[0]]; 111 for (IT = a[p].begin(); IT != a[p].end(); ++IT) { 112 seg_modify(seg[p], 1, n << 1, tr[*IT].st, 1); 113 seg_modify(seg[p], 1, n << 1, tr[*IT].ed, -1); 114 } 115 for (x = first[p]; x; x = e[x].next) 116 if (y != tr[p].fa[0]) dfs2(y); 117 } 118 #undef y 119 120 inline int get_lca(int x, int y) { 121 int i; 122 if (tr[x].dep < tr[y].dep) swap(x, y); 123 for (i = 17; ~i; --i) 124 if (tr[tr[x].fa[i]].dep >= tr[y].dep) 125 x = tr[x].fa[i]; 126 for (i = 17; ~i; --i) 127 if (tr[x].fa[i] != tr[y].fa[i]) 128 x = tr[x].fa[i], y = tr[y].fa[i]; 129 return x == y ? x : tr[x].fa[0]; 130 } 131 132 int main() { 133 int i, j, x, y, lca; 134 n = read(), m = read(); 135 for (i = 1; i < n; ++i) Add_edges(read(), read()); 136 for (i = 1; i <= m; ++i) { 137 x = read(), y = read(); 138 a[x].push_back(y); 139 Q[i] = Query(x, y); 140 } 141 seg[0] = ++cnt_seg, seg[0] -> ls = seg[0] -> rs = seg[0]; 142 tr[1].dep = 1; 143 dfs1(1); 144 dfs2(1); 145 for (i = 1; i <= m; ++i) { 146 x = Q[i].x, y = Q[i].y, lca = get_lca(x, y); 147 ans1 += get_ans(seg[x], seg[y], seg[lca], seg[tr[lca].fa[0]], 1, n << 1, tr[lca].st, tr[x].st); 148 ans1 += get_ans(seg[x], seg[y], seg[lca], seg[tr[lca].fa[0]], 1, n << 1, tr[lca].st, tr[y].st); 149 ans1 -= get_ans(seg[x], seg[y], seg[lca], seg[tr[lca].fa[0]], 1, n << 1, tr[lca].st, tr[lca].st); 150 --ans1; 151 } 152 sort(Q + 1, Q + m + 1); 153 for (i = 1; i <= m; i = j) { 154 for (j = i + 1; j <= m && Q[i] == Q[j]; ++j); 155 ans1 -= (ll) (j - i) * (j - i - 1) >> 1; 156 } 157 ans2 = (ll) m * (m - 1) >> 1; 158 g = gcd(ans1, ans2); 159 printf("%lld/%lld ", ans1 / g, ans2 / g); 160 return 0; 161 }