USACO Building Roads

洛谷 P2872 [USACO07DEC]道路建设Building Roads

洛谷传送门

JDOJ 2546: USACO 2007 Dec Silver 2.Building Roads

JDOJ传送门

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_i and 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

HINT

INPUT DETAILS:

Four farms at locations (1,1), (3,1), (2,3), and (4,3). Farms 1 and 4 are

connected by a road.

OUTPUT DETAILS:

Connect farms 1 and 2 with a road that is 2.00 units long, then connect

farms 3 and 4 with a road that is 2.00 units long. This is the best we can

do, and gives us a total of 4.00 unit lengths.

Source

2007~2008

题目翻译：

Farmer John最近得到了一些新的农场，他想新修一些道路使得他的所有农场可以经过原有的或是新修的道路互达（也就是说，从任一个农场都可以经过一些首尾相连道路到达剩下的所有农场）。有些农场之间原本就有道路相连。 所有N(1 <= N <= 1,000)个农场（用1..N顺次编号）在地图上都表示为坐标为(X_i, Y_i)的点(0 <= X_i <= 1,000,000；0 <= Y_i <= 1,000,000)，两个农场间道路的长度自然就是代表它们的点之间的距离。现在Farmer John也告诉了你农场间原有的M(1 <= M <= 1,000)条路分别连接了哪两个农场，他希望你计算一下，为了使得所有农场连通，他所需建造道路的最小总长是多少。

题解：

一道最小生成树的题。

这题的大意就是，有一些已经建好的边，问你再建多长的边能够使原图有一个最小生成树。

如果我们先依照题意连边未免太恶心。

所以我们考虑更优秀的一种做法（其实是更巧妙

我们先给题目的边打标记，最后建全图，如果有标记的边我们把它的边权置成0，这样就保证它一定在最小生成树上。

所以就可以A了。

（我用的是KUSKAL）

代码：

#include<cstdio>
#include<algorithm>
#include<cmath>
#define ll long long
using namespace std;
int n,m,vis[1001],tot,fa[1001];
long double ans;
struct node
{
    ll x,y;
}a[1001];
struct edge
{
    int u,v;
    long double val;
}e[1000001<<1];
long double dist(int x,int y,int a,int b)
{
    return sqrt((long double)(x-a)*(long double)(x-a)+(long double)(y-b)*(long double)(y-b));
}
void add(int x,int y,int z)
{
    e[++tot].u=x;
    e[tot].v=y;
    e[tot].val=z;
}
bool cmp(edge a,edge b)
{
    if(a.val==b.val)
        return a.u<b.u;
    return a.val<b.val;
}
int find(int x)
{
    if(fa[x]==x)
        return x;
    return fa[x]=find(fa[x]);
}
void unionn(int x,int y)
{
    int fx=find(x);
    int fy=find(y);
    if(fx!=fy)
        fa[fx]=fy;
}
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
        scanf("%d%d",&a[i].x,&a[i].y);
    for(int i=1;i<=n;i++)
        fa[i]=i;
    for(int i=1;i<=m;i++)
    {
        int u,v;
        scanf("%d%d",&u,&v);
        vis[u]=vis[v]=1;
    }
    for(int i=1;i<=n;i++)
        for(int j=i+1;j<=n;j++)
        {
            if(vis[i]==1 && vis[j]==1)
            {
                add(i,j,0);
                add(j,i,0);
            }
            else
            {
                add(i,j,dist(a[i].x,a[i].y,a[j].x,a[j].y));
                add(j,i,dist(a[i].x,a[i].y,a[j].x,a[j].y));
            }
        }
    sort(e+1,e+tot+1,cmp);
    for(int i=1;i<=tot;i++)
        if(find(e[i].u)!=find(e[i].v))
        {
            ans+=e[i].val;
            unionn(e[i].u,e[i].v);
        }
    printf("%.2lf",ans);
    return 0;
}