对于当天的营业额,你要找到之前的数和他差的绝对值最小的和。由于这个是不断在插入的,所以用伸展树来维护。

http://www.lydsy.com/JudgeOnline/problem.php?id=1588

照着kuangbin大牛的代码敲下来的。

#include<map>
#include<set>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define INF 99999999
#define ll __int64
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
const int MAXN = ;
int pre[MAXN],key[MAXN],ch[MAXN][],root,tot;
//父亲结点 键值 左右子树 根节点 节点数量
int n;
void Newnode(int &rt,int pa,int k)
{
rt = ++tot;
pre[rt] = pa;
key[rt] = k;
ch[rt][] = ch[rt][] = ;//空
} void Rotate(int x,int kind)
{
int y = pre[x];
ch[y][!kind] = ch[x][kind];
pre[ch[x][kind]] = y;
if(pre[y]){
ch[pre[y]][ch[pre[y]][]==y] = x;
}
pre[x] = pre[y];
ch[x][kind] = y;
pre[y] = x;
} //将结点rt 调整到goal下面
void splay(int rt,int goal)
{
while(pre[rt] != goal)
{
if(pre[pre[rt]] == goal)
Rotate(rt,ch[pre[rt]][]==rt);
else {
int y = pre[rt];
int kind = (ch[pre[y]][]==y);
if(ch[y][kind] == rt){//rt 和 pre[rt]不在同一方向上
Rotate(rt,!kind); //kind=1时,说明rt在y的右边 而y在pre[y]的左边
Rotate(rt,kind);
}
else {
Rotate(y,kind);
Rotate(rt,kind);
}
}
}
if(goal == )
root = rt;
} int Insert(int k)
{
int rt = root;
while(ch[rt][key[rt]<k]){//如果key[r]<k,表明当前k在右边,不然去左边
if(key[rt] == k){
splay(rt,);
return ;
}
rt = ch[rt][key[rt]<k];
}
Newnode(ch[rt][key[rt]<k],rt,k);
splay(ch[rt][key[rt]<k],);
return ;
} int get_pre(int rt)
{
int tmp = ch[rt][];
if(tmp == )
return INF;
while(ch[tmp][])
{
tmp = ch[tmp][];
}
return key[rt] - key[tmp];
} int get_next(int rt)
{
int tmp = ch[rt][];
if(tmp == )
return INF;
while(ch[tmp][])
{
tmp = ch[tmp][];
}
return key[tmp] - key[rt];
}
int main()
{
int i,j;
while(~scanf("%d",&n))
{
root = tot = ;
int ans = ;
for(i=; i<=n; i++){
int num;
if(scanf("%d",&num)==EOF)num = ;
if(i == ){
ans += num;
Newnode(root,,num);
continue;
}
if(Insert(num) == )
continue;
int a = get_pre(root);
int b = get_next(root);
//cout<<a<<" "<<b<<endl;
ans += min(a,b);
}
printf("%d\n",ans);
}
}
04-20 12:21