SPOJ:COT2 Count on a tree II
2018-04-06 07:57
411 查看
题意
给定一个n个节点的树,每个节点表示一个整数,问u到v的路径上有多少个不同的整数。n=40000,m=100000
Sol
树上莫队模板题# include <bits/stdc++.h> # define RG register # define IL inline # define Fill(a, b) memset(a, b, sizeof(a)) using namespace std; const int _(1e5 + 5); typedef long long ll; IL int Input(){ RG int x = 0, z = 1; RG char c = getchar(); for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1; for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48); return x * z; } int n, m, first[_], cnt, w[_], o[_], len; int dfn[_], st[20][_], lg[_], deep[_], idx, fa[_]; int S[_], bl[_], blo, num, ans[_], sum[_], vis[_], Ans; struct Edge{ int to, next; } edge[_]; struct Query{ int l, r, id; IL int operator <(RG Query B) const{ return bl[l] == bl[B.l] ? dfn[r] < dfn[B.r] : bl[l] < bl[B.l]; } } qry[_]; IL void Add(RG int u, RG int v){ edge[cnt] = (Edge){v, first[u]}, first[u] = cnt++; } IL void Dfs(RG int u){ dfn[u] = ++idx, st[0][idx] = u; RG int l = S[0]; for(RG int e = first[u]; e != -1; e = edge[e].next){ RG int v = edge[e].to; if(dfn[v]) continue; deep[v] = deep[u] + 1, fa[v] = u; Dfs(v); if(S[0] - l >= blo) for(++num; S[0] != l; --S[0]) bl[S[S[0]]] = num; st[0][++idx] = u; } S[++S[0]] = u; } IL void Chk(RG int &x, RG int u, RG int v){ x = deep[u] < deep[v] ? u : v; } IL int LCA(RG int u, RG int v){ u = dfn[u], v = dfn[v]; if(u > v) swap(u, v); RG int log2 = lg[v - u + 1], t; Chk(t, st[log2][u], st[log2][v - (1 << log2) + 1]); return t; } IL void Update(RG int x){ if(vis[x]) --sum[w[x]], Ans -= (!sum[w[x]]); else Ans += (!sum[w[x]]), ++sum[w[x]]; vis[x] ^= 1; } IL void Modify(RG int u, RG int v){ while(u != v){ if(deep[u] > deep[v]) swap(u, v); Update(v), v = fa[v]; } } int main(RG int argc, RG char* argv[]){ len = n = Input(), m = Input(), blo = sqrt(n); for(RG int i = 1; i <= n; ++i) o[i] = w[i] = Input(), first[i] = -1; sort(o + 1, o + len + 1), len = unique(o + 1, o + len + 1) - o - 1; for(RG int i = 1; i <= n; ++i) w[i] = lower_bound(o + 1, o + len + 1, w[i]) - o; for(RG int i = 1, u, v; i < n; ++i) u = Input(), v = Input(), Add(u, v), Add(v, u); Dfs(1); if(S[0]) for(++num; S[0]; --S[0]) bl[S[S[0]]] = num; for(RG int i = 2; i <= idx; ++i) lg[i] = lg[i >> 1] + 1; for(RG int j = 1; j <= lg[idx]; ++j) for(RG int i = 1; i + (1 << j) - 1 <= idx; ++i) Chk(st[j][i], st[j - 1][i], st[j - 1][i + (1 << (j - 1))]); for(RG int i = 1; i <= m; ++i){ qry[i] = (Query){Input(), Input(), i}; if(dfn[qry[i].l] > dfn[qry[i].r]) swap(qry[i].l, qry[i].r); } sort(qry + 1, qry + m + 1); RG int lca = LCA(qry[1].l, qry[1].r); Modify(qry[1].l, qry[1].r); Update(lca), ans[qry[1].id] = Ans, Update(lca); for(RG int i = 2; i <= m; ++i){ Modify(qry[i - 1].l, qry[i].l), Modify(qry[i - 1].r, qry[i].r); lca = LCA(qry[i].l, qry[i].r); Update(lca), ans[qry[i].id] = Ans, Update(lca); } for(RG int i = 1; i <= m; ++i) printf("%d\n", ans[i]); return 0; }
相关文章推荐
- spojCOT2 Count on a tree II
- [树上莫队] SPOJ COT2 Count on a tree II
- SPOJ-COT2 Count on a tree II(树上莫队)
- spoj_cot2 Count on a tree II(树上莫队+离散化)
- spoj_cot2 Count on a tree II(树上莫队+离散化)
- SPOJ COT2 Count on a tree II(路径树上莫队)
- SPOJ COT2 Count on a tree II 树上莫队
- [spoj COT2- Count on a tree II] 树上莫队
- 【SPOJ COT2】Count on a tree II,树上莫队
- [SPOJ COT2]-Count on a tree II-树上莫队
- AC日记——Count on a tree II spoj
- bzoj2589: Spoj 10707 Count on a tree II
- [BZOJ]2589: Spoj 10707 Count on a tree II
- BZOJ 2589 Spoj 10707 Count on a tree II 强制在线莫队算法(TLE)
- 【spoj】【COT2 - Count on a tree II】【莫队算法】
- 【SPOJ10707】Count on a tree II-树上莫队算法
- SPOJ 10707 Count on a Tree II 树上莫队
- SPOJ--Count on a tree II(树上莫队)
- 【SPOJ】Count On A Tree II(树上莫队)
- SPOJ10707 COT2 - Count on a tree II 【树上莫队】