题目大意

你被要求编写一个数据结构，支援以下操作，操作在线。

1. 插入一个元素
2. 查询一个元素与之前插入元素的最小差值。

题解

一道模板题。我是写了一个pre和succ函数水过的。1A，比较高兴。

代码

#include <algorithm>
#include <cstdio>
const int maxn = 40000;
int ch[maxn][2], fa[maxn], data[maxn], cnt[maxn], rt, sz, size[maxn];
void update(int k) { size[k] = size[ch[k][0]] + size[ch[k][1]] + cnt[k]; }
void zig(int x) {
int y = fa[x], z = fa[y], l = (ch[y][1] == x), r = l ^ 1;
fa[ch[y][l] = ch[x][r]] = y;
fa[ch[x][r] = y] = x;
fa[x] = z;
if (z)
ch[z][ch[z][1] == y] = x;
update(y);
}
void splay(int x, int aim = 0) {
for (int y; (y = fa[x]) != aim; zig(x))
if (fa[y] != aim)
zig((ch[y][0] == x) == (ch[fa[y]][0] == x) ? y : x);
if (aim == 0)
rt = x;
update(x);
}
void insert(int v) {
int x = rt;
if (rt == 0) {
rt = x = ++sz;
data[x] = v;
size[x] = cnt[x] = 1;
fa[x] = ch[x][0] = ch[x][1] = 0;
return;
}
while (x) {
size[x]++;
if (v == data[x]) {
cnt[x]++;
return;
}
int &y = ch[x][v >= data[x]];
if (y == 0) {
y = ++sz;
data[y] = v;
size[y] = cnt[y] = 1;
fa[y] = x;
ch[y][0] = ch[y][1] = 0;
x = y;
break;
}
x = y;
}
splay(x);
}
int pre(int v) {
int ans = -1;
for (int y = rt; y;) {
if (data[y] <= v)
ans = data[y], y = ch[y][1];
else
y = ch[y][0];
}
return ans;
}
int succ(int v) {
int ans = -1;
for (int y = rt; y;) {
if (data[y] >= v)
ans = data[y], y = ch[y][0];
else
y = ch[y][1];
}
return ans;
}
int main() {
#ifdef D
freopen("input", "r", stdin);
#endif
int ans = 0;
int n, x;
rt = 0;
scanf("%d %d", &n, &x);
insert(x);
ans += x;
for (int i = 2; i <= n; i++) {
scanf("%d", &x);
int ret = 0x3f3f3f;
int y = pre(x);
if (y != -1)
ret = std::min(ret, abs(x - y));
y = succ(x);
if (y != -1)
ret = std::min(ret, abs(x - y));
if (ret != 0x3f3f3f)
ans += ret;
insert(x);
}
printf("%d\n", ans);
return 0;
}