#include<bits/stdc++.h>
#define ll long long
using namespace std;
static const int N = 2e5+5;
static inline ll rd
() {
ll x = 0;
char d = getchar_unlocked
();
while
(d < '0' || d > '9'
)d = getchar_unlocked
();
while
(d >= '0' && d <= '9'
) {
x =
(x << 3
) +
(x << 1
) +
(d ^ 48
);
d = getchar_unlocked
();
}
return x;
}
static inline void wr
(ll x
) {
static ll len;
static char ch[20];
if
(x == 0
) {
putchar_unlocked
('0'
);
putchar_unlocked
(' '
);
return;
}
while
(x
) {
ch[len++] = x % 10 ^ 48;
x /= 10;
}
while
(len
)putchar_unlocked
(ch[--len]
);
putchar_unlocked
(' '
);
}
//------------------------------------------------
struct node {
ll x, y, id;
node
() {};
node
(ll a, ll b
): x
(a
), y
(b
) {};
} e;
static vector<node> v1, v2, v3, v4;
static inline bool p1
(const node &a, const node &b
) {
if
(a.x + a.y != b.x + b.y
)
return a.x + a.y < b.x + b.y;
if
(a.x != b.x
)
return a.x < b.x;
return a.y < b.y;
}
static inline bool p2
(const node &a, const node &b
) {
if
(a.x - a.y != b.x - b.y
)
return a.x - a.y < b.x - b.y;
if
(a.x != b.x
)
return a.x < b.x;
return a.y < b.y;
}
static inline bool p3
(const node &a, const node &b
) {
if
(-a.x + a.y != -b.x + b.y
)
return -a.x + a.y < -b.x + b.y;
if
(a.x != b.x
)
return a.x < b.x;
return a.y < b.y;
}
static inline bool p4
(const node &a, const node &b
) {
if
(-
(a.x + a.y
) != -
(b.x + b.y
))
return -
(a.x + a.y
) < -
(b.x + b.y
);
if
(a.x != b.x
)
return a.x < b.x;
return a.y < b.y;
}
struct edge {
ll u, v, w;
edge
() {};
edge
(ll a, ll b, ll c
): u
(a
), v
(b
), w
(c
) {};
bool operator <
(const edge& oth
)const {
return w < oth.w;
}
};
static inline ll get
(const node &a, const node &b
) {
return abs
(a.x - b.x
) + abs
(a.y - b.y
);
}
vector<edge> v;
vector<pair<int, int>> ans;
static int Set[N];
static inline int f
(int x
) {
return Set[x] == x ? x : Set[x] = f
(Set[x]
);
}
int main
() {
ios::sync_with_stdio
(0
);
cin.tie
(0
), cout.tie
(0
);
static ll n, sum = 0;
n = rd
();
v1.reserve
(n
);
v2.reserve
(n
);
v3.reserve
(n
);
v4.reserve
(n
);
v.reserve
(4 * n
);
for
(int i = 1; i <= n; i++
) {
e.x = rd
(), e.y = rd
(), e.id = i;
v1.emplace_back
(e
);
v2.emplace_back
(e
);
v3.emplace_back
(e
);
v4.emplace_back
(e
);
Set[i] = i;
}
//四向排序
sort
(v1.begin
(), v1.end
(), p1
);
sort
(v2.begin
(), v2.end
(), p2
);
sort
(v3.begin
(), v3.end
(), p3
);
sort
(v4.begin
(), v4.end
(), p4
);
//生成边
for
(int i = 1; i < n; i++
) {
v.emplace_back
(v1[i - 1].id, v1[i].id, get
(v1[i - 1], v1[i]
));
v.emplace_back
(v2[i - 1].id, v2[i].id, get
(v2[i - 1], v2[i]
));
v.emplace_back
(v3[i - 1].id, v3[i].id, get
(v3[i - 1], v3[i]
));
v.emplace_back
(v4[i - 1].id, v4[i].id, get
(v4[i - 1], v4[i]
));
}
sort
(v.begin
(), v.end
());
//kruskal
int Sz = v.size
(), rx = 0, ry = 0, cnt = 0;
for
(int i = 0; i < Sz; i++
) {
rx = f
(v[i].u
), ry = f
(v[i].v
);
if
(rx != ry
) {
sum += v[i].w;
Set[rx] = ry;
cnt++;
ans.emplace_back
(v[i].u, v[i].v
);
if
(cnt == n - 1
)break;
}
}
wr
(sum
);
for
(int i = 0; i<cnt; i++
) {
putchar_unlocked
('\n'
);
wr
(ans[i].first
), wr
(ans[i].second
);
}
return 0;
}还有那些问题？
鸡毛蒜皮就不用了
人们往往在小事上表现得更为理智，如购物时精挑细选；而在重大决定面前，比如选择居住城市或恋爱对象时，则可能更加冲动。本文通过一个个人故事探讨了这种行为模式。
扫一扫
188
被折叠的 条评论
为什么被折叠?
到【灌水乐园】发言
