标签:处理 algo 一个 for nbsp pair insert pos def
代码巨长的倍增题...
显然这是没有什么决策的,选择方案都是固定的
这可以考虑倍增跳,每个位置上跳到的位置是可以通过查前驱后继解决的
有两种方式:
一种是平衡树搞法,倒着做查完了插入
另一种是先排序建一个链表,查完了删除
都是可以通过加哨兵节点来搞的,
结果我只想到了 set 乱搞,就写了很长
预处理完就可做了
第一问对于每个点倍增一遍,其实就是照题意模拟,倍增优化一下
第二问还是照题意模拟,倍增优化一下
暴力有 70pts ?
代码:
#include <algorithm> #include <iostream> #include <cstdlib> #include <cstring> #include <cctype> #include <cstdio> #include <locale> #include <cmath> #include <set> using namespace std; typedef long long ll; const int MAX_N = 100005; const double INF = 100000000000001.0; struct Node { int len, id; Node(int L = 0, int ID = 0) {len = L; id = ID;} bool operator < (const Node& b) const { return len < b.len; } }; struct PAIR { ll fir, sec; PAIR(ll x = 0, ll y = 0) {fir = x; sec = y;} friend PAIR operator + (PAIR a, PAIR b) { return PAIR(a.fir + b.fir, a.sec + b.sec); } }d[MAX_N][20][2]; int n, lg, x0, m, ans; int h[MAX_N], f[MAX_N][20][2]; double ans_rat; set<Node> st; inline int rd() { register int x = 0, c = getchar(); register bool f = false; while (!isdigit(c)) { f = (c == ‘-‘); c = getchar(); } while (isdigit(c)) { x = x * 10 + (c ^ 48); c = getchar(); } return f ? -x : x; } inline void get_for(int pos, set<Node>::iterator iter) { auto lef = iter, rig = iter; if (iter == st.begin()) { f[pos][0][0] = iter->id; d[pos][0][0].fir = abs(h[pos] - h[iter->id]); ++iter; if (st.size() == 1) return; f[pos][0][1] = iter->id; d[pos][0][1].sec = abs(h[pos] - h[iter->id]); return; } else if (iter == st.end()) { --iter; f[pos][0][0] = iter->id; d[pos][0][0].fir = abs(h[pos] - h[iter->id]); if (iter == st.begin()) return; --iter; f[pos][0][1] = iter->id; d[pos][0][1].sec = abs(h[pos] - h[iter->id]); return; } ++rig; if (rig == st.end()) { --lef; if (lef == st.begin()) { if (abs(h[pos] - h[lef->id]) <= abs(h[pos] - h[iter->id])) { f[pos][0][0] = lef->id; d[pos][0][0].fir = abs(h[pos] - h[lef->id]); f[pos][0][1] = iter->id; d[pos][0][1].sec = abs(h[pos] - h[iter->id]); } else { f[pos][0][0] = iter->id; d[pos][0][0].fir = abs(h[pos] - h[iter->id]); f[pos][0][1] = lef->id; d[pos][0][1].sec = abs(h[pos] - h[lef->id]); } return; } if (abs(h[pos] - h[lef->id]) <= abs(h[pos] - h[iter->id])) { f[pos][0][0] = lef->id; d[pos][0][0].fir = abs(h[pos] - h[lef->id]); --lef; if (abs(h[pos] - h[lef->id]) <= abs(h[pos] - h[iter->id])) { f[pos][0][1] = lef->id; d[pos][0][1].sec = abs(h[pos] - h[lef->id]); } else { f[pos][0][1] = iter->id; d[pos][0][1].sec = abs(h[pos] - h[iter->id]); } return; } f[pos][0][0] = iter->id; d[pos][0][0].fir = abs(h[pos] - h[iter->id]); f[pos][0][1] = lef->id; d[pos][0][1].sec = abs(h[pos] - h[lef->id]); return; } --lef; if (abs(h[pos] - h[lef->id]) == abs(h[pos] - h[iter->id])) { f[pos][0][0] = lef->id; d[pos][0][0].fir = abs(h[pos] - h[lef->id]); f[pos][0][1] = iter->id; d[pos][0][1].sec = abs(h[pos] - h[iter->id]); } else if (abs(h[pos] - h[lef->id]) < abs(h[pos] - h[iter->id])) { f[pos][0][0] = lef->id; d[pos][0][0].fir = abs(h[pos] - h[lef->id]); if (lef == st.begin()) { f[pos][0][1] = iter->id; d[pos][0][1].sec = abs(h[pos] - h[iter->id]); } else { --lef; if (abs(h[pos] - h[lef->id]) <= abs(h[pos] - h[iter->id])) { f[pos][0][1] = lef->id; d[pos][0][1].sec = abs(h[pos] - h[lef->id]); } else { f[pos][0][1] = iter->id; d[pos][0][1].sec = abs(h[pos] - h[iter->id]); } } } else { f[pos][0][0] = iter->id; d[pos][0][0].fir = abs(h[pos] - h[iter->id]); if (abs(h[pos] - h[lef->id]) <= abs(h[pos] - h[rig->id])) { f[pos][0][1] = lef->id; d[pos][0][1].sec = abs(h[pos] - h[lef->id]); } else { f[pos][0][1] = rig->id; d[pos][0][1].sec = abs(h[pos] - h[rig->id]); } } } inline void DBL_init() { for (int i = n; i >= 1; --i) { if (!st.empty()) { auto dst = st.lower_bound(h[i]); get_for(i, dst); } st.insert(Node(h[i], i)); } for (int i = 1; i <= n && 2 + i <= n; ++i) { f[i][1][0] = f[f[i][0][0]][0][1]; f[i][1][1] = f[f[i][0][1]][0][0]; d[i][1][0] = d[i][0][0] + d[f[i][0][0]][0][1]; d[i][1][1] = d[i][0][1] + d[f[i][0][1]][0][0]; } for (int j = 2; j <= lg; ++j) { for (int i = 1; i <= n && (1 << j) + i <= n; ++i) { f[i][j][0] = f[f[i][j - 1][0]][j - 1][0]; f[i][j][1] = f[f[i][j - 1][1]][j - 1][1]; d[i][j][0] = d[i][j - 1][0] + d[f[i][j - 1][0]][j - 1][0]; d[i][j][1] = d[i][j - 1][1] + d[f[i][j - 1][1]][j - 1][1]; } } } inline void get_ans(int bgn) { register int pos = bgn; register ll dst_a = 0, dst_b = 0; register bool got_bgn = false; register double rat; for (int i = lg; i >= 0; --i) { if (!f[pos][i][1] || dst_a + dst_b + d[pos][i][1].fir + d[pos][i][1].sec > x0 || (i == 0 && !f[pos][0][0])) continue; got_bgn = true; dst_b += d[pos][i][1].fir; dst_a += d[pos][i][1].sec; pos = f[pos][i][1]; } if (!got_bgn) return; if (dst_b == 0) { if (dst_a == 0) rat = 1.0; else rat = INF; } else { rat = double(dst_a) / double(dst_b); } if (fabs(0.0 - ans_rat) < 1e-7 || rat < ans_rat || (fabs(rat - ans_rat) < 1e-7 && h[bgn] > h[ans])) { ans_rat = rat; ans = bgn; } } int main() { n = rd(); lg = int(ceil(log2(n))); for (int i = 1; i <= n; ++i) h[i] = rd(); DBL_init(); x0 = rd(); for (int i = 1; i <= n; ++i) get_ans(i); printf("%d\n", ans); m = rd(); register int s = 0, x = 0; register ll tot_a = 0, tot_b = 0; while (m--) { tot_a = tot_b = 0; s = rd(); x = rd(); for (int i = lg; i >= 0; --i) { if (!f[s][i][1] || (tot_a + tot_b + d[s][i][1].fir + d[s][i][1].sec > x)) continue; tot_b += d[s][i][1].fir; tot_a += d[s][i][1].sec; s = f[s][i][1]; } printf("%lld %lld\n", tot_a, tot_b); } return 0; }
标签:处理 algo 一个 for nbsp pair insert pos def
原文地址:https://www.cnblogs.com/xcysblog/p/9868559.html