poj 3625 Building Roads

题目连接

Building Roads

Description

Farmer John had just acquired several new farms! He wants to connect the farms with roads so that he can travel from any farm to any other farm via a sequence of roads; roads already connect some of the farms.

Each of the N (1 ≤ N ≤ 1,000) farms (conveniently numbered 1..N) is represented by a position (X_i, Y_i) on the plane (0 ≤ X_i≤ 1,000,000; 0 ≤ Y_i≤ 1,000,000). Given the preexisting M roads (1 ≤ M ≤ 1,000) as pairs of connected farms, help Farmer John determine the smallest length of additional roads he must build to connect all his farms.

Input

* Line 1: Two space-separated integers: N and M
* Lines 2..N+1: Two space-separated integers: X_iand Y_i
* Lines N+2..N+M+2: Two space-separated integers: i and j, indicating that there is already a road connecting the farm i and farm j.

Output

* Line 1: Smallest length of additional roads required to connect all farms, printed without rounding to two decimal places. Be sure to calculate distances as 64-bit floating point numbers.

Sample Input

4 1
1 1
3 1
2 3
4 3
1 4

Sample Output

4.00

最小生成树。。

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<cmath>
#include<set>
using std::set;
using std::sort;
using std::pair;
using std::swap;
using std::multiset;
#define pb(e) push_back(e)
#define sz(c) (int)(c).size()
#define mp(a, b) make_pair(a, b)
#define all(c) (c).begin(), (c).end()
#define iter(c) decltype((c).begin())
#define cls(arr, val) memset(arr, val, sizeof(arr))
#define cpresent(c, e) (find(all(c), (e)) != (c).end())
#define rep(i, n) for(int i = 0; i < (int)n; i++)
#define tr(c, i) for(iter(c) i = (c).begin(); i != (c).end(); ++i)
const int N = 1010;
const int INF = 0x3f3f3f3f;
typedef unsigned long long ull;
struct Node {
	int x, y;
	double w;
	Node() {}
	Node(int i, int j, double k) :x(i), y(j), w(k) {}
	inline bool operator<(const Node &t) const {
		return w < t.w;
	}
}G[(N * N) << 1];
struct P {
	double x, y;
	inline double calc(const P &t) const {
		return sqrt((x - t.x) * (x - t.x) + (y - t.y) * (y - t.y));
	}
}A[N];
struct Kruskal {
	int E, par[N], rank[N];
	inline void init() {
		E = 0;
		rep(i, N) {
			par[i] = i;
			rank[i] = 0;
		}
	}
	inline int find(int x) {
		while (x != par[x]) {
			x = par[x] = par[par[x]];
		}
		return x;
	}
	inline bool unite(int x, int y) {
		x = find(x), y = find(y);
		if (x == y) return false;
		if (rank[x] < rank[y]) {
			par[x] = y;
		} else {
			par[y] = x;
			rank[x] += rank[x] == rank[y];
		}
		return true;
	}
	inline void built(int n, int m) {
		int u, v;
		for(int i = 1; i<= n; i++) scanf("%lf %lf", &A[i].x, &A[i].y);
		for (int i = 1; i <= n; i++) {
			for (int j = i + 1; j <= n; j++) {
				G[E++] = Node(i, j, A[i].calc(A[j]));
			}
		}
		while (m--) {
			scanf("%d %d", &u, &v);
			G[E++] = Node(u, v, 0.0);
		}
	}
	inline double kruskal(int n) {
		int tot = 0;
		double ans = 0.0;
		sort(G, G + E);
		rep(i, E) {
			Node &e = G[i];
			if (unite(e.x, e.y)) {
				ans += e.w;
				if (++tot >= n - 1) return ans;
			}
		}
		return -1.0;
	}
	inline void solve(int n, int m) {
		init(), built(n, m);
		printf("%.2lf
", kruskal(n));
	}
}go;
int main() {
#ifdef LOCAL
	freopen("in.txt", "r", stdin);
	freopen("out.txt", "w+", stdout);
#endif
	int n, m;
	while (~scanf("%d %d", &n, &m)) {
		go.solve(n, m);
	}
	return 0;
}