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 (Xi, Yi) on the plane (0 ≤ Xi ≤ 1,000,000; 0 ≤ Yi ≤ 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: Xi and Yi
* 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\n", 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;
}
poj 3625 Building Roads的更多相关文章
- poj 3625 Building Roads(最小生成树,二维坐标,基础)
题目 //最小生成树,只是变成二维的了 #define _CRT_SECURE_NO_WARNINGS #include<stdlib.h> #include<stdio.h> ...
- HDU 1815, POJ 2749 Building roads(2-sat)
HDU 1815, POJ 2749 Building roads pid=1815" target="_blank" style="">题目链 ...
- poj 2749 Building roads (二分+拆点+2-sat)
Building roads Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 6229 Accepted: 2093 De ...
- [poj] 2749 building roads
原题 2-SAT+二分答案! 最小的最大值,这肯定是二分答案.而我们要2-SATcheck是否在该情况下有可行解. 对于目前的答案limit,首先把爱和恨连边,然后我们n^2枚举每两个点通过判断距离来 ...
- POJ 2749 Building roads 2-sat+二分答案
把爱恨和最大距离视为限制条件,可以知道,最大距离和限制条件多少具有单调性 所以可以二分最大距离,加边+check #include<cstdio> #include<algorith ...
- Building roads
Building roads Time Limit: 10000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Tot ...
- [POJ2749]Building roads(2-SAT)
Building roads Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 8153 Accepted: 2772 De ...
- poj 1251 Jungle Roads (最小生成树)
poj 1251 Jungle Roads (最小生成树) Link: http://poj.org/problem?id=1251 Jungle Roads Time Limit: 1000 ...
- 多次访问节点的DFS POJ 3411 Paid Roads
POJ 3411 Paid Roads Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 6553 Accepted: 24 ...
随机推荐
- IE SEESION共享的问题
前几天,我们在开发工作流的过程中出现了一个比较奇怪的问题,原本看不到流程的人员,在登陆后却能够看到对应流程的待办任务,并且导致流程流向混乱!在调模式下调试程序发现(假设登陆两个用户)第二个登陆用户的信 ...
- JS的文本编辑框jwysiwyg-0.6
一款轻量的用js写的文本编辑框.
- .NET本质论(4)应用程序对象HttpApplication
当HttpContext对象创建之后,HttpRuntime将随后创建一个用于处理请求的对象,这个对象的类型为HttpApplication. 在ASP.NET内部,HttpRuntime管理一个定 ...
- centos7 搭建docker内运行rabbitmq,然后再镜像ha方案的完全教程,暂时一个宿主机只能运行一个docker的rabbitmq,但是集群 ha都正常
1.安装centos7.x,配置好网络2.因为docker需要比较高版本的内核,比如使用overlayfs作为默认docker文件系统要3.18,所以先升级内核到3.18以上版本,能直接过4是最佳了检 ...
- Sunglasses
It's hot this summer. It also reminds me of one case about sunglasses. She was new to this company a ...
- 右下角弹出"Windows-延缓写入失败"或者"xxx-损坏文件 请运行Chkdsk工具"
知识点分析: 任务栏右下角弹出“Windows-延缓写入失败”或者“xxx-损坏文件 请运行Chkdsk工具”. 操作步骤: 方法一:Chkdsk工具 在开始---运行中输入cmd,然后输入chkds ...
- Android IOS WebRTC 音视频开发总结(二一)-- 黑屏问题
本文主要介绍音视频通话中收到第一帧图像后视频一直卡住的问题,文章来自博客园RTC.Blacker,转载请说明出处. 因为苹果AppStore要求从2015年2月1日开始所有所有上架App必须支持arm ...
- Xcode Alcatraz插件管理介绍和使用
Xcode Alcatraz插件管理介绍和使用http://www.jianshu.com/p/7a2484123bf6 1.简介 Alcatraz是一个能帮你管理Xcode插件丶模版及颜色配置的工具 ...
- 使用OrderBy对List<Person>集合排序
string sortOrder = Request.QueryString["sortOrder"]; string sortField = Request.QueryStr ...
- jquery Ajax中的dataType简析
jquery中的ajax有好几种运用方式,但是基本上都是使用的$.ajax()方法,很多人经常会使用它来从后台获取json格式的数据,但是经常发现返回的json字符串并不能自动的转换成js里的json ...