标签:amp 恶心 nta 元素 def 时间复杂度 iostream i++ bsp
由于困难值小于等于x这个很恶心,可以离线处理,将边权,和询问时的x排序。
每到一个询问的时候,将边权小于等于x的都合并起来再询问。
。。
有重复元素的线段树合并的时间复杂度是nlog^2n
#include <cstdio> #include <iostream> #include <algorithm> #define N 500001 int n, m, q, cnt, tot, size; int sum[N * 10], ls[N * 10], rs[N * 10], a[N], b[N], f[N], root[N], ans[N], c[N << 1]; struct node { int x, y, z, id; node(int x = 0, int y = 0, int z = 0, int id = 0) : x(x), y(y), z(z), id(id) {} }p[N], ask[N]; inline int read() { int x = 0, f = 1; char ch = getchar(); for(; !isdigit(ch); ch = getchar()) if(ch == ‘-‘) f = -1; for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - ‘0‘; return x * f; } inline bool cmp1(node x, node y) { return x.z < y.z; } inline bool cmp2(node x, node y) { return x.y < y.y; } inline void merge(int &x, int y) { if(!x || !y) { x += y; return; } sum[x] += sum[y]; merge(ls[x], ls[y]); merge(rs[x], rs[y]); } inline void insert(int &now, int l, int r, int x) { now = ++size; if(l == r) { sum[now] = 1; return; } int mid = (l + r) >> 1; if(x <= mid) insert(ls[now], l, mid, x); else insert(rs[now], mid + 1, r, x); sum[now] = sum[ls[now]] + sum[rs[now]]; } inline int query(int now, int l, int r, int x) { if(l == r) return l; int mid = (l + r) >> 1; if(x <= sum[ls[now]]) return query(ls[now], l, mid, x); else return query(rs[now], mid + 1, r, x - sum[ls[now]]); } inline int find(int x) { return x == f[x] ? x : f[x] = find(f[x]); } int main() { int i, j, x, y, z; n = read(); m = read(); q = read(); for(i = 1; i <= n; i++) a[i] = b[i] = read(); std::sort(b + 1, b + n + 1); cnt = std::unique(b + 1, b + n + 1) - b - 1; for(i = 1; i <= n; i++) { a[i] = std::lower_bound(b + 1, b + cnt + 1, a[i]) - b; f[i] = i; insert(root[i], 1, cnt, a[i]); } for(i = 1; i <= m; i++) { x = read(); y = read(); c[i] = z = read(); p[i] = node(x, y, z, 0); } for(i = 1; i <= q; i++) { x = read(); c[i + m] = y = read(); z = read(); ask[i] = node(x, y, z, i); } std::sort(c + 1, c + m + q + 1); tot = std::unique(c + 1, c + m + q + 1) - c - 1; for(i = 1; i <= m; i++) p[i].z = std::lower_bound(c + 1, c + tot + 1, p[i].z) - c; for(i = 1; i <= q; i++) ask[i].y = std::lower_bound(c + 1, c + tot + 1, ask[i].y) - c; std::sort(p + 1, p + m + 1, cmp1); std::sort(ask + 1, ask + q + 1, cmp2); j = 1; for(i = 1; i <= q; i++) { while(j <= m && p[j].z <= ask[i].y) { x = find(p[j].x); y = find(p[j].y); if(x ^ y) { f[y] = x; merge(root[x], root[y]); } j++; } x = find(ask[i].x); if(ask[i].z > sum[root[x]]) ans[ask[i].id] = -1; else ans[ask[i].id] = b[query(root[x], 1, cnt, sum[root[x]] - ask[i].z + 1)]; } for(i = 1; i <= q; i++) printf("%d\n", ans[i]); return 0; }
[BZOJ3545] [ONTAK2010]Peaks(线段树合并 + 离散化)
标签:amp 恶心 nta 元素 def 时间复杂度 iostream i++ bsp
原文地址:http://www.cnblogs.com/zhenghaotian/p/7518804.html