AtCoder Regular Contest 077 D Built?[最小生成树]

传送门：http://arc076.contest.atcoder.jp/tasks/arc076_b

Sample Input 1

3
1 5
3 9
7 8

Sample Output 1

3

Sample Input 2

6
8 3
4 9
12 19
18 1
13 5
7 6

Sample Output 2

8

题意：求1e5个点的最小生成树，每两个点a，b距离为min(|Xa-Xb|,|Ya-Yb|)。  kruskal求最小生成树 点数较多 考虑存在三个点a b c满足Xa<Xb<Xc 连接a和c花费为Xc-Xa，相比连接a，b和b，c 价格相同，所以后者连接了三个点更优

所以只需要分别排序后处理相邻点的2*(N-1)条边，复杂度nlogn。

代码：

#define _CRT_SECURE_NO_DEPRECATE
#pragma comment(linker, "/STACK:102400000,102400000")
#include<iostream>  
#include<cstdio>  
#include<fstream>  
#include<iomanip>
#include<algorithm>  
#include<cmath>  
#include<deque>  
#include<vector>  
#include<assert.h>
#include<bitset>
#include<queue>  
#include<string>  
#include<cstring>  
#include<map>  
#include<stack>  
#include<set>
#include<functional>
#define pii pair<int, int>
#define mod 1000000007
#define mp make_pair
#define pi acos(-1)
#define eps 0.00000001
#define mst(a,i) memset(a,i,sizeof(a))
#define all(n) n.begin(),n.end()
#define lson(x) ((x<<1))  
#define rson(x) ((x<<1)|1) 
#define inf 0x3f3f3f3f
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
const int maxn = 1e5 + 5;
class node {
public:
    int u, v, w;
    node(int a, int b, int c) :u(a), v(b), w(c) {};
};
int fa[maxn];
int getf(int i) { return i == fa[i] ? i : fa[i] = getf(fa[i]); }
vector<node>edge;
bool cmp(const node&a, const node&b){return a.w < b.w;}
class T {
public:
    int value, id;
    bool operator<(const T &a){return value < a.value;}
};
T sx[maxn], sy[maxn];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int i, j, k, m, n, ta, tb;
    cin >> n;
    for (int i = 1; i <= n; ++i)
        fa[i] = i;
    for (int i = 1; i <= n; ++i)
    {
        cin >> sx[i].value >> sy[i].value;
        sx[i].id = sy[i].id = i;
    }
    sort(sx + 1, sx + n + 1);
    sort(sy + 1, sy + 1 + n);
    for (int i = 1; i < n; ++i)
    {
        edge.push_back(node(sx[i].id, sx[i + 1].id, sx[i+1].value - sx[i].value));
        edge.push_back(node(sy[i].id, sy[i + 1].id, sy[i+1].value - sy[i].value));
    }
    sort(all(edge), cmp);
    ll ans = 0;
    for (int i = 0; i <edge.size(); ++i)
    {
        if (getf(edge[i].u) != getf(edge[i].v))
        {
            ans += edge[i].w;
            fa[getf(edge[i].u)] = getf(edge[i].v);
        }
    }
    cout << ans << endl;
    return 0;
}