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CodeForces - 343D 树链剖分
题目链接:http://codeforces.com/problemset/problem/343/D
题意:给定一棵n个n-1条边的树,起初所有节点权值为0,然后m个操作。 1 x:把x为根的子树的点的权值修改为1; 2 x:把x结点到根路径上的点修改为0; 3 x:查询结点x的值。
思路:树链剖分。 对于操作1,子树操作记录下当前点为根时,dfs的最后一个点的id是多少(endid[]),然后就可以把子树操作用区间来维护了。 对于操作2,树链剖分。 对于3操作,线段树单点查询。
#define _CRT_SECURE_NO_DEPRECATE #include<iostream> #include<cstring> #include<string> #include<algorithm> #include<stdio.h> #include<queue> #include<vector> #include<stack> #include<map> #include<set> #include<time.h> #include<cmath> #include<sstream> #include<assert.h> using namespace std; #define L(x) x<<1 #define R(x) x<<1|1 typedef long long int LL; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3fLL; const int MAXN = 500000 + 10; int head[MAXN], tot, cnt; struct Edge{ int to, next; }Edges[MAXN * 2]; void add(int u, int v){ Edges[tot].to = v; Edges[tot].next = head[u]; head[u] = tot++; } int id[MAXN], endid[MAXN],son[MAXN], deep[MAXN], size[MAXN], fa[MAXN], reid[MAXN], top[MAXN]; void Init(){ tot = 0; cnt = 0; memset(head, -1, sizeof(head)); memset(son, -1, sizeof(son)); } void DFS1(int u, int p,int dep){ fa[u] = p; size[u] = 1; deep[u] = dep; for (int i = head[u]; i != -1; i = Edges[i].next){ if (Edges[i].to != p){ DFS1(Edges[i].to, u,dep+1); size[u] += size[Edges[i].to]; if (son[u] == -1 || size[Edges[i].to] > size[son[u]]){ son[u] = Edges[i].to; } } } } void DFS2(int u, int tp){ id[u] = ++cnt; reid[id[u]] = u; top[u] = tp; if (son[u] == -1){ endid[u] = cnt; return; } DFS2(son[u], tp); for (int i = head[u]; i != -1; i = Edges[i].next){ if (son[u] != Edges[i].to&&Edges[i].to != fa[u]){ DFS2(Edges[i].to, Edges[i].to); } } endid[u] = cnt; } struct Node{ int st, ed; int lazy,val; }Seg[MAXN * 4]; void pushDown(int k){ if (Seg[k].lazy!=-1){ Seg[L(k)].lazy = Seg[L(k)].val = Seg[k].lazy; Seg[R(k)].lazy = Seg[R(k)].val = Seg[k].lazy; Seg[k].lazy = -1; } } void Build(int l, int r, int k){ Seg[k].st = l; Seg[k].ed = r; Seg[k].lazy = -1; Seg[k].val = 0; if (l == r){ return; } int mid = (l + r) / 2; Build(l, mid, L(k)); Build(mid + 1, r, R(k)); } void Modify(int l,int r,int val,int k){ if (Seg[k].st == l&&Seg[k].ed == r){ Seg[k].lazy = Seg[k].val = val; return; } pushDown(k); if (r <= Seg[L(k)].ed){ Modify(l, r, val, L(k)); } else if (l >= Seg[R(k)].st){ Modify(l, r, val, R(k)); } else{ Modify(l, Seg[L(k)].ed, val, L(k)); Modify(Seg[R(k)].st, r, val, R(k)); } } void Modify(int u, int v,int val){ int f1 = top[u], f2 = top[v]; while (f1 != f2){ if (deep[f1] < deep[f2]){ swap(f1, f2); swap(u, v); } Modify(id[f1], id[u], val,1); u = fa[f1]; f1 = top[u]; } if (deep[u] > deep[v]){ swap(u, v); } Modify(id[u], id[v],val, 1); } int Query(int pos, int k){ if (Seg[k].st == Seg[k].ed){ return Seg[k].val; } pushDown(k); if (pos <= Seg[L(k)].ed){ return Query(pos, L(k)); } else{ return Query(pos, R(k)); } } int main(){ //#ifdef kirito // freopen("in.txt", "r", stdin); // freopen("out.txt", "w", stdout); //#endif // int start = clock(); int n,m; while (~scanf("%d",&n)){ Init(); for (int i = 1; i < n; i++){ int u, v; scanf("%d%d", &u, &v); add(u, v); add(v, u); } DFS1(1, 1, 0); DFS2(1, 1); Build(1, n, 1); scanf("%d", &m); while (m--){ int u, v; scanf("%d%d", &u, &v); switch (u){ case 1: Modify(id[v], endid[v], 1, 1); break; case 2: Modify(v, 1, 0); break; default: printf("%d\n", Query(id[v], 1)); break; } } } //#ifdef LOCAL_TIME // cout << "[Finished in " << clock() - start << " ms]" << endl; //#endif return 0; }
CodeForces - 343D 树链剖分
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