【wikioi】1296 营业额统计
题目链接:http://www.wikioi.com/problem/1296/
算法:Splay
这是非常经典的一道题目,用Splay树来维护营业额,每天的最小波动值就等于 min{树根-树根的前驱, 树根的后继-树根)
所以用Splay来维护
PS: 本题数据有问题,所以当空行时,值为0
==========================14.06.13==========================
原来写的splay的bug太多,已换成数组= = ps:14.07.26又换成指针。。。。>_<
详细看另一篇splay文章,http://www.cnblogs.com/iwtwiioi/p/3537061.html
=================================很久以前===============================
代码:
#include <cstdio>
using namespace std;
#define F(rt) rt-> pa
#define K(rt) rt-> key
#define CH(rt, d) rt-> ch[d]
#define C(rt, d) (K(rt) > d ? 0 : 1)
#define NEW(d) new Splay(d)
#define PRE(rt) F(rt) = CH(rt, 0) = CH(rt, 1) = null int n, ans; struct Splay {
Splay* ch[2], *pa;
int key;
Splay(int d = 0) : key(d) { ch[0] = ch[1] = pa = NULL; }
}; typedef Splay* tree;
tree null = new Splay, root = null; void rot(tree& rt, int d) {
tree k = CH(rt, d^1), u = F(rt); int flag = CH(u, 1) == rt;
CH(rt, d^1) = CH(k, d); if(CH(k, d) != null) F(CH(k, d)) = rt;
CH(k, d) = rt; F(rt) = k; rt = k; F(rt) = u;
if(u != null) CH(u, flag) = k;
} void splay(tree nod, tree& rt) {
if(nod == null) return;
tree pa = F(rt);
while(F(nod) != pa) {
if(F(nod) == rt)
rot(rt, CH(rt, 0) == nod);
else {
int d = CH(F(F(nod)), 0) == F(nod);
int d2 = CH(F(nod), 0) == nod;
if(d == d2) { rot(F(F(nod)), d); rot(F(nod), d2); }
else { rot(F(nod), d2); rot(F(nod), d); }
}
}
rt = nod;
} tree maxmin(tree rt, int d) {
if(rt == null) return null;
while(CH(rt, d) != null) rt = CH(rt, d);
return rt;
} tree ps(tree rt, int d) {
if(rt == null) return null;
rt = CH(rt, d);
return maxmin(rt, d^1);
} tree search(tree& rt, int d) {
tree t = rt;
while(t != null && K(t) != d) t = CH(t, C(t, d));
splay(t, rt);
return t;
} void insert(tree& rt, int d) {
tree q = NULL, t = rt;
while(t != null) q = t, t = CH(t, C(t, d));
t = NEW(d);
PRE(t);
if(q) F(t) = q, CH(q, C(q, d)) = t;
else rt = t;
splay(t, rt);
} void del(tree& rt) {
if(rt == null) return;
tree t = rt;
if(CH(t, 0) == null) t = CH(rt, 1);
else {
t = CH(rt, 0);
splay(maxmin(t, 1), t);
CH(t, 1) = CH(rt, 1);
if(CH(rt, 1) != null) F(CH(rt, 1)) = t;
}
delete rt;
F(t) = null;
rt = t;
} void init(int key) {
if(root == null) { root = NEW(key); PRE(root); ans += key; return; }
insert(root, key);
tree succ = ps(root, 0), pred = ps(root, 1);
if(succ == null) { ans += K(pred) - K(root); splay(pred, root); return; }
if(pred == null) { ans += K(root) - K(succ); splay(succ, root); return; }
int l = K(root) - K(succ), r = K(pred) - K(root);
if(l <= r) { ans += l; splay(succ, root); }
else { ans += r; splay(pred, root); }
} int main() {
PRE(null);
scanf("%d", &n);
int c;
for(int i = 0; i < n; ++i) { if(scanf("%d", &c) == EOF) c = 0; init(c); } //坑爹的读入
printf("%d\n", ans);
return 0;
}
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