题目链接
题意
有$n$个人，每个人都有两个属性，现在要选出若干个人使用第一属性，其余人使用第二属性。使第一属性最大值和第二属性的最大值差的绝对值最小。
思路
很容易想到枚举每一个人作为第一属性的能力值最高的人，那么那些第一属性比他高的人都去发挥第二属性，我们维护这批人的第二属性最大值（因为这批人只会往里面添加，不会减少）。我们还可以在其余人里选出一些发挥第二属性，很容易发现选第二属性接近我们当前枚举的人的第一属性的人。
那么实现上就只需要一个能维护插入元素的前驱和后继并快速删除的数据结构。
如，treap，pbds，splay等。
牢牢记住这些东西删除重复元素是全部删除的！！！1
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using namespace std;
typedef long long ll;
void err(){cout << "\033[39;0m" << endl;}
template<template<typename...> class T, typename t, typename... A>
void err(T<t> a, A... x){for (auto v: a) cout << v << ' '; err(x...);}
template<typename T, typename... A>
void err(T a, A... x){cout << a << ' '; err(x...);}
const int N = 1e5 + 5;
namespace Treap
{
struct node
{
int child[2], size;
ll val;
unsigned key;
}treap[N];
/*
ll seed = 17124057;
ll Rand()
{
return seed = seed * 48271 % 2147483647;
}
*/
mt19937 rd(time(0));
unsigned Rand[20];
void rand_init()
{
for (int i = 0; i < 20; i++)
Rand[i] = rd();
}
int cur;
inline void init(int &rt)
{
rt = cur = 0;
rand_init();
ls(0) = rs(0) = 0;
treap[0].size = 0;
treap[0].val = 0;
}
void dfs(int root)
{
dbg(root, ls(root), rs(root));
if (ls(root))
dfs(ls(root));
if (rs(root))
dfs(rs(root));
}
inline void pushup(int root)
{
treap[root].size = treap[ls(root)].size + treap[rs(root)].size + 1;
}
void split(int root, int &x, int &y, ll v)
{
if (!root)
{
x = 0;
y = 0;
return;
}
if (treap[root].val <= v)
{
x = root;
split(rs(root), rs(x), y, v);
}
else
{
y = root;
split(ls(root), x, ls(y), v);
}
pushup(root);
}
void merge(int &root, int x, int y)
{
if (!x || !y)
{
root = x + y;
return;
}
if (treap[x].key <= treap[y].key)
{
root = x;
merge(rs(root), rs(x), y);
}
else
{
root = y;
merge(ls(root), x, ls(y));
}
pushup(root);
}
void split_by_rank(int root, int &x, int &y, int k)
{
//dbg(root, k);
if (!root)
{
x = y = 0;
return;
}
if (treap[ls(root)].size + 1 <= k)
{
x = root;
split_by_rank(rs(root), rs(x), y, k - treap[ls(root)].size - 1);
}
else
{
y = root;
split_by_rank(ls(root), x, ls(y), k);
}
pushup(root);
}
inline unsigned get_rd(int id)
{
unsigned now = 0;
for (int i = 0; i < 20; i++)
if (id & (1 << i))
now ^= Rand[i];
return now;
}
inline void insert(int &root, ll v)
{
int x = 0, y = 0, z = ++cur;
treap[z].val = v;
ls(z) = rs(z) = 0;
treap[z].size = 1;
treap[z].key = get_rd(cur);
split(root, x, y, v);
merge(x, x, z);
merge(root, x, y);
}
inline void del(int &root, ll v)
{
//dbg("del which val is ", v);
int x = 0, y = 0, z = 0;
split(root, x, y, v);
int rk = treap[x].size;
split_by_rank(x, x, z, rk - 1);
//dbg(z, treap[z].size, treap[z].val);
assert(treap[z].size == 1);
//dfs(x);
//dfs(y);
//dfs(z);
merge(root, x, y);
}
inline pair<int, int> lower_bound(int &root, ll v)
{
//dbg("ask near to ", v);
int x = 0, y = 0;
split(root, x, y, v - 1);
int now = y;
while (ls(now))
now = ls(now);
int nnow = x;
while (rs(nnow))
nnow = rs(nnow);
merge(root, x, y);
// dbg(nnow, now, treap[nnow].val, treap[now].val);
return make_pair(nnow, now);
}
int rt;
} //namespace Treap
struct Node
{
ll x, y;
}arr[N];
bool cmp(Node x, Node y)
{
if (x.x == y.x)
return x.y < y.y;
return x.x > y.x;
}
/*
int main()
{
Treap::init(Treap::rt);
Treap::insert(Treap::rt, 14);
Treap::insert(Treap::rt, 87);
Treap::insert(Treap::rt, 9);
Treap::insert(Treap::rt, 47);
Treap::insert(Treap::rt, 36);
Treap::dfs(Treap::rt);
Treap::del(Treap::rt, 14);
Treap::del(Treap::rt, 87);
return 0;
}
*/
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int n;
scanf("%d", &n);
for (int i = 1; i <= n; i++)
{
scanf("%lld%lld", &arr[i].x, &arr[i].y);
arr[i].x += 1;
arr[i].y += 1;
}
sort(arr + 1, arr + n + 1, cmp);
Treap::init(Treap::rt);
ll maxi = -5e18;
for (int i = 1; i <= n; i++)
{
Treap::insert(Treap::rt, arr[i].y);
//dbg(i, arr[i].x, arr[i].y);
}
ll ans = 5e18;
for (int i = 1; i <= n; i++)
{
Treap::del(Treap::rt, arr[i].y);
if (maxi >= arr[i].x)
{
ans = min(ans, maxi - arr[i].x);
}
else{
pair<int, int> p = Treap::lower_bound(Treap::rt, arr[i].x);
int pre = p.first, suf = p.second;
//dbg(pre, suf, Treap::treap[pre].val, Treap::treap[suf].val, maxi);
if (suf)
ans = min(ans, abs(max(maxi, Treap::treap[suf].val) - arr[i].x));
if (pre)
{
ans = min(ans, abs(max(maxi, Treap::treap[pre].val) - arr[i].x));
}
//dbg(ans);
ans=min(ans,abs(maxi-arr[i].x));
}
maxi=max(maxi,arr[i].y);
}
printf("%lld\n",ans);
}
return 0;
}