贪心思想。将a排序后，对于每一个a，找到对应的删除m个后最小的b，每次更新答案即可。
如何删除才是合法并且最优的？首先，对于排了序的a，第$i$个那么之前就应该删除前$i-1$个a对应的b。剩下$m-i+1$可以删，那么在剩下的b中查找第$m-i+2$小即可。每次做完就删除当前a对应的b。
注意离散化。
还有数组不要开大了....
#include<bits/stdc++.h> #define LL long long using namespace std; void read(int &x) { x = 0; int t = 1; char ch = getchar(); while(ch > '9' || ch < '0') { if(ch == '-') t = -1; ch = getchar(); } while(ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); } x = x * t; } int a[500005], b[500005], n, m; struct Node { int a, b; int ida, idb; } mat[100005]; bool cmp(Node a, Node b) { if(a.a == b.a) return a.b < b.b; return a.a < b.a; } int siz[400005], val[400005]; void modify(int nd, int l, int r, int pos, int d) { if(l == r) { siz[nd] += d; return ; } int mid = (l + r) >> 1; if(pos <= mid) modify(nd << 1, l, mid, pos, d); else modify(nd << 1 | 1, mid + 1, r, pos, d); siz[nd] = siz[nd << 1] + siz[nd << 1 | 1]; } int query(int nd, int l, int r, int pos) { if(l == r) return val[l]; int mid = (l + r) >> 1; if(pos <= siz[nd << 1]) return query(nd << 1, l, mid, pos); else return query(nd << 1 | 1, mid + 1, r, pos - siz[nd << 1]); } int pos[100005]; void work() { sort(b + 1, b + 1 + n); int q = unique(b + 1, b + 1 + n) - b - 1; memset(siz, 0, sizeof(siz)); sort(mat + 1, mat + 1 + n, cmp); for(int i = 1; i <= n; i ++) { pos[i] = lower_bound(b + 1, b + 1 + q, mat[i].b) - b; val[pos[i]] = mat[i].b; modify(1, 1, q, pos[i], 1); } LL ans = 0; int r = min(mat[1].b, query(1, 1, q, m + 1)); ans = 1ll * mat[1].a * r; for(int i = 1; i <= m; i ++) { modify(1, 1, q, pos[i], -1); int res = m - i + 1; r = query(1, 1, q, res); if(mat[i + 1].b < r) r = mat[i + 1].b; LL now = 1ll * mat[i + 1].a * r; ans = max(ans, now); } printf("%lld\n", ans); } int main() { freopen("d.in", "r", stdin); freopen("d.out", "w", stdout); int T; scanf("%d", &T); while(T --) { read(n); read(m); for(int i = 1; i <= n; i ++) { read(mat[i].a); read(mat[i].b); b[i] = mat[i].b; } work(); } return 0; }
依旧是数据结构。我们发现，对于一个点集，它们一定有一个lca，而满足包含所有点的最小点集就是所有点到这个lca的链上的所有点。再加上题目要求最小的$abs(a[u]-r)$，就是用每条链上r的前驱和后驱来更新答案。
用主席树维护即可。每个结点的版本是由它的父亲结点转移过来。这样做查询的时候直接取出当前结点和lca的父亲结点的版本即可。
关于这个前驱和后驱的计算可以学习一下。
#include<bits/stdc++.h> #define LL long long using namespace std; const int maxn = 1e9; void read(int &x) { x = 0; int t = 1; char ch = getchar(); while(ch > '9' || ch < '0') { if(ch == '-') t = -1; ch = getchar(); } while(ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); } x = x * t; } int n, q, type; struct Node { int v, nex; Node(int v = 0, int nex = 0) : v(v), nex(nex) { } } Edge[200005]; int h[100005], stot; void add(int u, int v) { Edge[++stot] = Node(v, h[u]); h[u] = stot; } int sum[32*100005], ls[32*100005], rs[32*100005], tail; inline int newnode(int x) { sum[++tail] = sum[x]; ls[tail] = ls[x]; rs[tail] = rs[x]; return tail; } inline void update(int nd) { sum[nd] = sum[ls[nd]] + sum[rs[nd]]; } int insert(int nd, int l, int r, LL pos) { nd = newnode(nd); sum[nd] ++; if(l == r) return nd; int mid = (l + r) >> 1; if(pos <= mid) ls[nd] = insert(ls[nd], l, mid, pos); else rs[nd] = insert(rs[nd], mid + 1, r, pos); update(nd); return nd; } int queryl(int nd1, int nd2, int l, int r, int pos) { if(sum[nd1] == sum[nd2]) return 0;///这个结点下面没有新的点 if(l == r) return l; int mid = (l + r) >> 1; if(pos <= mid) return queryl(ls[nd1], ls[nd2], l, mid, pos); else { int tmp = queryl(rs[nd1], rs[nd2], mid + 1, r, pos);////尽量往靠近pos的地方走 if(tmp) return tmp;///如果走不到就尽量往mid走 else return queryl(ls[nd1], ls[nd2], l, mid, mid); } } int queryr(int nd1, int nd2, int l, int r, int pos) { if(sum[nd1] == sum[nd2]) return 0; if(l == r) return l; int mid = (l + r) >> 1; if(pos > mid) return queryr(rs[nd1], rs[nd2], mid + 1, r, pos); else { int tmp = queryr(ls[nd1], ls[nd2], l, mid, pos); if(tmp) return tmp; else return queryr(rs[nd1], rs[nd2], mid + 1, r, mid); } } int dep[100005], jum[100005][21]; int lca(int u, int v) { if(dep[u] < dep[v]) swap(u, v); int t = dep[u] - dep[v]; for(int p = 0; t; t >>= 1, p ++) if(t & 1) u = jum[u][p]; if(u == v) return u; for(int p = 19; p >= 0; p --) if(jum[u][p] != jum[v][p]) u = jum[u][p], v = jum[v][p]; return jum[u][0]; } int rt[100005], a[100005]; void dfs(int u, int f) { jum[u][0] = f; for(int i = 1; i < 20; i ++) jum[u][i] = jum[jum[u][i - 1]][i - 1]; dep[u] = dep[f] + 1; rt[u] = insert(rt[f], 1, maxn, a[u]); for(int i = h[u]; i; i = Edge[i].nex) { int v = Edge[i].v; if(v == f) continue; dfs(v, u); } } int x[100005], p[100005]; int main() { freopen("e.in", "r", stdin); freopen("e.out", "w", stdout); scanf("%d%d%d", &n, &q, &type); for(int i = 1; i <= n; i ++) read(a[i]); for(int i = 1; i < n; i ++) { int u, v; read(u); read(v); add(u, v); add(v, u); } dfs(1, 0); int lastans = 0; for(int i = 1; i <= q; i ++) { int r, k; scanf("%d%d", &r, &k); for(int j = 1; j <= k; j ++) { read(x[j]); p[j] = (x[j] - 1 + lastans * type) % n + 1; } int l = p[1]; for(int j = 2; j <= k; j ++) l = lca(l, p[j]); l = jum[l][0]; int res = maxn, tmp; for(int j = 1; j <= k; j ++) { tmp = queryl(rt[l], rt[p[j]], 1, maxn, r); if(tmp && r - tmp < res) res = r - tmp; tmp = queryr(rt[l], rt[p[j]], 1, maxn, r); if(tmp && tmp - r < res) res = tmp - r; } printf("%d\n", res); lastans = res; } return 0; }