题目描述

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.

给定 nnn 个点的坐标，第 iii 个点的坐标为 (xi,yi)，这 n 个点编号为 1 到 n。给定 m 条边，第 i 条边连接第 ui 个点和第 vi 个点。现在要求你添加一些边，并且能使得任意一点都可以连通其他所有点。求添加的边的总长度的最小值。

输入格式

* 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.

第一行两个整数 n,m 代表点数与边数。
接下来 n 行每行两个整数 xi,yi 代表第 i 个点的坐标。
接下来 m 行每行两个整数 ui,vi 代表第 iii 条边连接第 ui 个点和第 vi 个点。

输出格式

* 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.

一行一个实数代表添加的边的最小长度，要求保留两位小数，为了避免误差，请用 64 位实型变量进行计算。

输入输出样例

输入 #1
4 1
1 1
3 1
2 3
4 3
1 4
输出 #1
4.00
说明/提示

数据规模与约定

说明

Translated by 一只书虫仔。

思路：

将已连的边的边权设置成 0

再将未连接的所有点之间的边连接起来（需要计算边权）

最后跑最小生成树，把其边权相加就是答案！

ps：这个题卡map，判断是否已连接的时候换成bool 数组来判断，用map会T两个测试点

AC代码如下：

#include <bits/stdc++.h>
#define buff                     \ios::sync_with_stdio(false); \cin.tie(0);                  \cout.tie(0)
using namespace std;
const int N = 1e3 + 9;
int n, m;
struct point
{double x, y;
} s[N];
double g[N][N];
bool st[N];
double dist[N];
//map<pair<int,int>,int> mp;//map的查找速度是log(n)级别
//.......卡map
bool mp[N][N];
void add(int a, int b, bool flag)
{if (flag){g[a][b] = 0;g[b][a] = 0;return ;}double d = sqrt((s[a].x - s[b].x) * (s[a].x - s[b].x) + (s[a].y - s[b].y) * (s[a].y - s[b].y));g[a][b] = d;g[b][a] = d;
}
void init()
{for(int i = 1;i<=n;i++){dist[i] = 0x3f3f3f3f;}
}
double prim()
{init();double res = 0;for (int i = 0; i < n; i++){int t = -1;for (int j = 1; j <= n; j++){if (!st[j] && (t == -1 || dist[t] > dist[j]))t = j;}if (i && dist[t] == 0x3f3f3f3f)return 0x3f3f3f3f;if (i)res += dist[t];st[t] = 1;for (int j = 1; j <= n; j++){dist[j] = min(dist[j], g[t][j]);}}return res;
}
int main()
{buff;cin >> n >> m;for (int i = 1; i <= n; i++){cin >> s[i].x >> s[i].y;}for (int i = 1; i <= m; i++){int a, b;cin >> a >> b;add(a, b, 1);// mp[{a,b}] = 1;// mp[{b,a}] = 1;mp[a][b] = 1;mp[b][a] = 1;}for(int i = 1;i<=n;i++)for(int j = 1;j<=n;j++){if(!mp[i][j]&&i!=j)add(i,j,0);}double ans = prim();printf("%.2lf\n", ans);
}

P2872 [USACO07DEC]Building Roads S（最小生成树）相关推荐

洛谷——P2872 [USACO07DEC]道路建设Building Roads
P2872 [USACO07DEC]道路建设Building Roads 题目描述 Farmer John had just acquired several new farms! He wants ...
USACO07DEC道路建设Building Roads（prim算法+堆优化与Kruskal+路径压缩对比）
目录 primprimprim算法 KruskalKruskalKruskal算法 P2872 [USACO07DEC]道路建设Building Roads 4 1 1 1 3 1 2 3 4 3 1 ...
bzoj 1626: [Usaco2007 Dec]Building Roads 修建道路（最小生成树）
1626: [Usaco2007 Dec]Building Roads 修建道路 Time Limit: 5 Sec Memory Limit: 64 MB Submit: 1709 Solved ...
[BZOJ1626][Usaco2007 Dec]Building Roads 修建道路
1626: [Usaco2007 Dec]Building Roads 修建道路 Time Limit: 5 Sec Memory Limit: 64 MB Submit: 1730 Solved ...
HDU 1301 Jungle Roads(裸最小生成树）
题目链接今天做了好几个模版最小生成树...贴一个kurskral. 1 /* 2 HDU 1301 Jungle Roads 3 最小生成树Kurskal模版 4 */ 5 #include < ...
HDU 1815， POJ 2749 Building roads（2-sat）
HDU 1815, POJ 2749 Building roads 题目链接HDU 题目链接POJ 题意: 有n个牛棚, 还有两个中转站S1和S2, S1和S2用一条路连接起来. 为了使得随意牛棚两个 ...
【HDU - 1102】Constructing Roads （最小生成树裸题模板）
题干: There are N villages, which are numbered from 1 to N, and you should build some roads such that ...
Building Roads（POJ-3625）
Problem Description Farmer John had just acquired several new farms! He wants to connect the farms w ...
HDOJ1102 Constructing Roads【最小生成树】-----武科大ACM暑期集训队选拔赛1题
这道题目没有做出来,代码写好之后一直没有AC,本以为做了这么多最小生成树的题目,这道题一定没问题的,结果很遗憾,没有注意细节问题: 首先,如何处理已经建好的路?已经建好的路说明这两个点是连通的,只要把 ...

P2872 [USACO07DEC]Building Roads S（最小生成树）

题目描述

输入格式

输出格式

输入输出样例

说明/提示

数据规模与约定

说明

思路：

AC代码如下：

P2872 [USACO07DEC]Building Roads S（最小生成树）相关推荐

最新文章

热门文章