multiset || 线段树 HDOJ 4302 Holedox Eating
题意:一个长度L的管子,起点在0。n次操作,0 p表示在p的位置放上蛋糕,1表示去吃掉最近的蛋糕(如果左右都有蛋糕且距离相同,那么吃同方向的蛋糕),问最终走了多少路程
分析:用multiset来保存蛋糕的位置,以当前的位置进行二分查找相邻的蛋糕的位置,模拟这个过程。当然也可以用线段树单点更新维护。
收获:当结果和暴力程序跑出来的一样却WA时,想想是否遗漏了某些情况的讨论
multiset:
/************************************************ * Author :Running_Time * Created Time :2015-8-22 9:31:50 * File Name :C.cpp ************************************************/ #include <cstdio> #include <algorithm> #include <iostream> #include <sstream> #include <cstring> #include <cmath> #include <string> #include <vector> #include <queue> #include <deque> #include <stack> #include <list> #include <map> #include <set> #include <bitset> #include <cstdlib> #include <ctime> using namespace std; #define lson l, mid, rt << 1 #define rson mid + 1, r, rt << 1 | 1 typedef long long ll; const int N = 1e5 + 10; const int INF = 0x3f3f3f3f; const int MOD = 1e9 + 7; int main(void) { int T, cas = 0; scanf ("%d", &T); while (T--) { multiset<int> S; S.clear (); multiset<int>::iterator it, l, r; int n, m; scanf ("%d%d", &n, &m); ll ans = 0; int now = 0, d = 1; //d == 0:左 d == 1: 右 while (m--) { int op, p; scanf ("%d", &op); if (op == 0) { scanf ("%d", &p); S.insert (p); } else { if (S.empty ()) continue; //没蛋糕了 it = S.lower_bound (now); r = it, l = it; if (it == S.end ()) { //右边没蛋糕 l--; ans += (now - *l); if (*l < now) d = 0; //当蛋糕在左边时才改变方向,下面类似 now = *l; S.erase (l); } else if (it == S.begin ()) { //左边没蛋糕 ans += (*r - now); if (now < *r) d = 1; now = *r; S.erase (r); } else { //左右都有蛋糕 l--; int dt0 = now - *l, dt1 = *r - now; if (dt0 == dt1) { //左右距离相等 if (d == 0) { ans += dt0; now = *l; S.erase (l); } else { ans += dt1; now = *r; S.erase (r); } } else if (dt0 < dt1) { //吃左边的蛋糕 ans += dt0; now = *l; if (dt0) d = 0; S.erase (l); } else { //吃右边的蛋糕 ans += dt1; now = *r; if (dt1) d = 1; S.erase (r); } } } } printf ("Case %d: %I64d\n", ++cas, ans); } return 0; }
Segment_Tree:
#include <bits/stdc++.h> using namespace std; #define lson l, mid, o << 1 #define rson mid + 1, r, o << 1 | 1 typedef long long ll; const int N = 1e5 + 5; const int INF = 0x3f3f3f3f; struct Segment_Tree { struct Node { int mx, mn, cnt; }node[N<<2]; void push_up(int o) { node[o].mx = max (node[o<<1].mx, node[o<<1|1].mx); node[o].mn = min (node[o<<1].mn, node[o<<1|1].mn); } void build(int l, int r, int o) { if (l == r) { node[o].mx = -INF; node[o].mn = INF; node[o].cnt = 0; return ; } int mid = l + r >> 1; build (lson); build (rson); push_up (o); } void updata(int p, int c, int l, int r, int o) { if (l == r && p == l) { node[o].mx = node[o].mn = p; node[o].cnt += c; if (!node[o].cnt) { node[o].mx = -INF; node[o].mn = INF; } return ; } int mid = l + r >> 1; if (p <= mid) updata (p, c, lson); else updata (p, c, rson); push_up (o); } int query_left(int ql, int qr, int l, int r, int o) { if (ql <= l && r <= qr) { return node[o].mx; } int mid = l + r >> 1, ret = -INF; if (ql <= mid) ret = max (ret, query_left (ql, qr, lson)); if (qr > mid) ret = max (ret, query_left (ql, qr, rson)); return ret; } int query_right(int ql, int qr, int l, int r, int o) { if (ql <= l && r <= qr) { return node[o].mn; } int mid = l + r >> 1, ret = INF; if (ql <= mid) ret = min (ret, query_right (ql, qr, lson)); if (qr > mid) ret = min (ret, query_right (ql, qr, rson)); return ret; } }st; int main(void) { int T, cas = 0; scanf ("%d", &T); while (T--) { int n, m; scanf ("%d%d", &n, &m); n++; st.build (1, n, 1); ll ans = 0; int now = 1, d = 1; while (m--) { int op, p; scanf ("%d", &op); if (op == 0) { scanf ("%d", &p); p++; st.updata (p, 1, 1, n, 1); } else { int p1 = st.query_left (0, now, 1, n, 1); int p2 = st.query_right (now, n, 1, n, 1); if (p1 == -INF && p2 == INF) continue; if (p2 == INF) { ans += now - p1; if (p1 < now) d = 0; now = p1; st.updata (p1, -1, 1, n, 1); } else if (p1 == -INF) { ans += p2 - now; if (p2 > now) d = 1; now = p2; st.updata (p2, -1, 1, n, 1); } else { int dt0 = now - p1, dt1 = p2 - now; if (dt0 == dt1) { if (d == 0) { ans += dt0; now = p1; st.updata (p1, -1, 1, n, 1); } else { ans += dt1; now = p2; st.updata (p2, -1, 1, n, 1); } } else if (dt0 < dt1) { ans += dt0; now = p1; if (dt0) d = 0; st.updata (p1, -1, 1, n, 1); } else { ans += dt1; now = p2; if (dt1) d = 1; st.updata (p2, -1, 1, n, 1); } } } } printf ("Case %d: %I64d\n", ++cas, ans); } return 0; }
编译人生,运行世界!