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:
    吴裕雄--天生自然C++语言学习笔记:C++ 标准库
    吴裕雄--天生自然C++语言学习笔记:C++ STL 教程
    吴裕雄--天生自然C++语言学习笔记:C++ Web 编程
    吴裕雄--天生自然C++语言学习笔记:C++ 多线程
    吴裕雄--天生自然C++语言学习笔记:C++ 信号处理
    吴裕雄--天生自然C++语言学习笔记:C++ 预处理器
    吴裕雄--天生自然C++语言学习笔记:C++ 模板
    吴裕雄--天生自然C++语言学习笔记:C++ 命名空间
    ZOJ1905Power Strings (KMP||后缀数组+RMQ求循环节)
    POJ3693Maximum repetition substring (循环节)(后缀数组+RMQ)
  • 原文地址:https://www.cnblogs.com/GadyPu/p/4782252.html
Copyright © 2011-2022 走看看