POJ 1751 Highways

題目大意：

        只有一個測例，有N個點編號1 ~ N（1 ≤ N ≤ 750），並給出N個點的座標"X Y"，其中0 ≤ X, Y ≤ 10,000，都是整型，接下來會給出M條已經存在的邊（0 ≤ M ≤ 1,000），每條邊都給出兩個端點的編號，問題是還要添加幾條邊才能使圖上所有點都連通並且是新添加的邊的總長度最小，只要求輸出添加的每條邊，每條邊包含兩個點的編號（用空格隔開），每條邊佔一行，每條邊中點的順序以及邊的輸出順序任意。

題目鏈接

註釋代碼：

/*                                    
 * Problem ID : POJ 1751 Highways 
 * Author     : Lirx.t.Una                                    
 * Language   : C++                        
 * Run Time   : 79 ms                                    
 * Run Memory : 2880 KB                                    
*/

#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>

//點的最大數量
#define	MAXN	750

using namespace std;

struct	Point {//點的位置信息

	int		x, y;

	friend istream &
	operator>>(istream &is, Point &p) {
	
		is >> p.x >> p.y;
		return is;
	}

	int
	POW(int a) { return a * a; }

	int
	operator^(Point &oth) {//計算兩點距離的平方
	
		return POW( x - oth.x ) + POW( y - oth.y );
	}
};

struct	Arc {//邊的信息，輸出答案時用到

	int		u, v;//邊的兩個端點

	Arc(int uu, int vv) : u(uu), v(vv) {}

	friend ostream &
	operator<<(ostream &os, Arc &arc) {
	
		os << arc.u << ' ' << arc.v;
		return os;
	}
};

struct	Node {//Prim結點

	int		u;
	int		d;

	Node(void) {}

	Node(int uu, int dd) : u(uu), d(dd) {}

	bool
	operator<(const Node &oth)
	const {

		return d > oth.d;
	}
};

Point	p[MAXN + 1];
int		g[MAXN + 1][MAXN + 1];

int		pre[MAXN + 1];
int		d[MAXN + 1];
bool	vis[MAXN + 1];

queue<Arc>		ans;
priority_queue<Node>	heap;

void
prim(int n) {

	int		i;
	int		nv;

	int		u, v;

	Node	node;

	vis[1] = true;
	for ( i = 2; i <= n; i++ ) {
	
		d[i]   = g[1][i];
		pre[i] = 1;

		heap.push(Node( i, d[i] ));
	}

	nv = 1;
	while ( !heap.empty() ) {//輸入會出現無法構成生成樹的情況！！蛋疼
	
		while (true) {
		
			node = heap.top();
			heap.pop();

			if ( !vis[ u = node.u ] ) {
			
				vis[u] = true;
				nv++;

				if ( g[ pre[u] ][u] ) ans.push(Arc(pre[u], u));//不是已經造好的路就放入ans隊列中
				break;
			}
		}

		if ( nv == n ) break;

		for ( v = 2; v <= n; v++ )
			if ( !vis[v] && g[u][v] < d[v] ) {

				d[v]   = g[u][v];
				pre[v] = u;

				heap.push(Node( v, d[v] ));
			}
	}

	while ( !ans.empty() ) {//將需要新建的路輸出
	
		cout << ans.front() << endl;
		ans.pop();
	}
}

int
main() {

	int		n, m;//點數，已經修好的路的數量
	int		u, v;//臨時點
	int		i, j;

	scanf("%d", &n);
	for ( i = 1; i <= n; i++ ) cin >> p[i];
	for ( i = 1; i <= n; i++ )
		for ( j = i + 1; j <= n; j++ )
			g[i][j] = g[j][i] = p[i] ^ p[j];

	scanf("%d", &m);
	while ( m-- ) {
	
		scanf("%d%d", &u, &v);
		g[u][v] = g[v][u] = 0;//已經修好的路就不需要耗費
	}

	prim(n);

	return 0;
}

無註釋代碼：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>

#define	MAXN	750

using namespace std;

struct	Point {

	int		x, y;

	friend istream &
	operator>>(istream &is, Point &p) {
	
		is >> p.x >> p.y;
		return is;
	}

	int
	POW(int a) { return a * a; }

	int
	operator^(Point &oth) {
	
		return POW( x - oth.x ) + POW( y - oth.y );
	}
};

struct	Arc {

	int		u, v;

	Arc(int uu, int vv) : u(uu), v(vv) {}

	friend ostream &
	operator<<(ostream &os, Arc &arc) {
	
		os << arc.u << ' ' << arc.v;
		return os;
	}
};

struct	Node {

	int		u;
	int		d;

	Node(void) {}

	Node(int uu, int dd) : u(uu), d(dd) {}

	bool
	operator<(const Node &oth)
	const {

		return d > oth.d;
	}
};

Point	p[MAXN + 1];
int		g[MAXN + 1][MAXN + 1];

int		pre[MAXN + 1];
int		d[MAXN + 1];
bool	vis[MAXN + 1];

queue<Arc>		ans;
priority_queue<Node>	heap;

void
prim(int n) {

	int		i;
	int		nv;

	int		u, v;

	Node	node;

	vis[1] = true;
	for ( i = 2; i <= n; i++ ) {
	
		d[i]   = g[1][i];
		pre[i] = 1;

		heap.push(Node( i, d[i] ));
	}

	nv = 1;
	while ( !heap.empty() ) {
	
		while (true) {
		
			node = heap.top();
			heap.pop();

			if ( !vis[ u = node.u ] ) {
			
				vis[u] = true;
				nv++;

				if ( g[ pre[u] ][u] ) ans.push(Arc(pre[u], u));
				break;
			}
		}

		if ( nv == n ) break;

		for ( v = 2; v <= n; v++ )
			if ( !vis[v] && g[u][v] < d[v] ) {

				d[v]   = g[u][v];
				pre[v] = u;

				heap.push(Node( v, d[v] ));
			}
	}

	while ( !ans.empty() ) {
	
		cout << ans.front() << endl;
		ans.pop();
	}
}

int
main() {

	int		n, m;
	int		u, v;
	int		i, j;

	scanf("%d", &n);
	for ( i = 1; i <= n; i++ ) cin >> p[i];
	for ( i = 1; i <= n; i++ )
		for ( j = i + 1; j <= n; j++ )
			g[i][j] = g[j][i] = p[i] ^ p[j];

	scanf("%d", &m);
	while ( m-- ) {
	
		scanf("%d%d", &u, &v);
		g[u][v] = g[v][u] = 0;
	}

	prim(n);

	return 0;
}