zoukankan      html  css  js  c++  java
  • poj 3625 Building Roads

    题目连接

    http://poj.org/problem?id=3625  

    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 (XiYi) on the plane (0 ≤ X≤ 1,000,000; 0 ≤ Y≤ 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: Xand Y
    * 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;
    }
  • 相关阅读:
    Celery
    mysql 8.0.12 创建并授权出现的问题
    request对象
    Haystack搜索框架
    Django的缓存机制
    跨域问题
    解析器
    url控制器与响应器
    学期总结
    C语言I博客作业09
  • 原文地址:https://www.cnblogs.com/GadyPu/p/4782252.html
Copyright © 2011-2022 走看看