SPOJ 10628 Count on a tree (lca+主席树)
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题意:给定一棵有n个结点的树,每一个点有一个权值。共同拥有m个询问。对于每一个询问(u,v,k),回答结点u至v之间第k小的点的权值。
思路:主席树+lca。首先指定一个根结点dfs一次并在此过程中建好主席树。对于对于每一个询问,我们仅仅须要考虑四棵树,即T[u], T[v], T[lca(u,v)], 再加上T[fa( lca(u,v) )],fa( lca(u,v) )表示lca(u, v)的父亲结点。
这样一来问题就和线性序列里第k小的数一样了。
#include<cstdio> #include<cstring> #include<cmath> #include<cstdlib> #include<iostream> #include<algorithm> #include<vector> #include<map> #include<queue> #include<stack> #include<string> #include<map> #include<set> #include<ctime> #define eps 1e-6 #define LL long long #define pii (pair<int, int>) //#pragma comment(linker, "/STACK:1024000000,1024000000") using namespace std; //const int maxn = 100000 + 100; //const int INF = 0x3f3f3f3f; const int maxn = 100000+10000; const int M = 2000000; int n, q, m, tot; int t[maxn], w[maxn], fa[maxn]; int T[maxn], lson[M], rson[M], c[M]; struct Quest { int l, r, k; } quest[maxn]; void Init_hash(int k) { sort(t, t+k); m = unique(t, t+k) - t; } int Hash(int x) { return lower_bound(t, t+m, x) - t; } int build(int l, int r) { int root = tot++; c[root] = 0; if(l != r) { int mid = (l+r) >> 1; lson[root] = build(l, mid); rson[root] = build(mid+1, r); } return root; } int Insert(int root, int pos, int val) { int newroot = tot++, tmp = newroot; int l = 0, r = m-1; c[newroot] = c[root] + val; while(l < r) { int mid = (l+r)>>1; if(pos <= mid) { lson[newroot] = tot++; rson[newroot] = rson[root]; newroot = lson[newroot]; root = lson[root]; r = mid; } else { rson[newroot] = tot++; lson[newroot] = lson[root]; newroot = rson[newroot]; root = rson[root]; l = mid+1; } c[newroot] = c[root] + val; } return tmp; } int Query(int l_root, int r_root, int lca, int k) { int l = 0, r = m -1, lca_root = T[lca], fa_root = fa[lca]; while(l < r) { int mid = (l+r) >> 1; int tmp = c[lson[l_root]]+c[lson[r_root]]-c[lson[lca_root]]-c[lson[fa_root]]; if(tmp >= k) { r = mid; l_root = lson[l_root]; r_root = lson[r_root]; lca_root = lson[lca_root]; fa_root = lson[fa_root]; } else { l = mid + 1; k -= tmp; l_root = rson[l_root]; r_root = rson[r_root]; lca_root = rson[lca_root]; fa_root = rson[fa_root]; } } return l; } int pnt[maxn], lca[maxn]; bool vis[maxn]; vector<int> G[maxn], query[maxn], num[maxn]; int find(int x) { if(x == pnt[x]) return x; return pnt[x] = find(pnt[x]); } void dfs_lca(int u) { vis[u] = 1; pnt[u] = u; int sz1 = G[u].size(); for(int i = 0; i < sz1; i++) { int v = G[u][i]; if(vis[v]) continue; fa[v] = T[u]; dfs_lca(v); pnt[v] = u; } int sz2 = query[u].size(); for(int i = 0; i < sz2; i++) { int v = query[u][i]; if(vis[v]) lca[num[u][i]] = find(v); } } void init() { memset(vis, 0, sizeof(vis)); for(int i = 1; i <= n; i++) { G[i].clear(); query[i].clear(); num[i].clear(); } } void dfs_ZXTree(int cur, int fa) { int sz = G[cur].size(); for(int i = 0; i < sz; i++) { int u = G[cur][i]; if(u == fa) continue; T[u] = Insert(T[cur], Hash(w[u]), 1); dfs_ZXTree(u, cur); } } int main() { //freopen("input.txt", "r", stdin); while(cin >> n >> q) { init(); m = 0; tot = 0; for(int i = 1; i <= n; i++) scanf("%d", &w[i]), t[m++] = w[i]; Init_hash(m); build(0, m-1); for(int i = 1; i < n; i++) { int u, v; scanf("%d%d", &u, &v); G[u].push_back(v); G[v].push_back(u); } T[1] = Insert(T[0], Hash(w[1]), 1); dfs_ZXTree(1, -1); for(int i = 0; i < q; i++) { int l, r, k; scanf("%d%d%d", &l, &r, &k); quest[i].l = l; quest[i].r = r; quest[i].k = k; query[l].push_back(r); query[r].push_back(l); num[l].push_back(i); num[r].push_back(i); } fa[1] = T[0]; dfs_lca(1); for(int i = 0; i < q; i++) { printf("%d\n", t[Query(T[quest[i].l], T[quest[i].r], lca[i], quest[i].k)]); } } return 0; }
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