#include<bits/std
c++.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
本文记录了一道C++在线判题（OJ）问题，涉及如何处理输入并输出。当遇到包含多个单词的输入行时，需要输出每一行，且逐行将第一个单词移到末尾。讲解了两种解决方案：一是结合`while`循环与`cin.get()`判断输入结束，二是使用`istringstream`处理输入。还讨论了C++中控制输入格式和未格式化的输入/输出操作。
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