poj 3625 Building Roads（最小生成树）

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 (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

代码（Kruskal算法）：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
using namespace std;

#define maxn 1000+10
#define maxv 1000000+10
struct edge
{
    int u,v;
    double w;
    friend bool operator < (edge a,edge b)
    {
        return a.w<b.w;
    }
} e[maxv];
int pre[maxn];
int x[maxn],y[maxn];
int n,m,len;

void init()
{
    for(int i=0; i<=n; i++)
        pre[i]=i;
}

int Find(int c)
{
    if(c==pre[c])
        return c;
    pre[c]=Find(pre[c]);
    return pre[c];
}

int join(int cx,int cy)
{
    int fx=Find(cx),fy=Find(cy);
    if(fx!=fy)
    {
        pre[fy]=fx;
        return 1;
    }
    return 0;
}

void Kruskal()
{
    sort(e,e+len);
    int cont=0;
    double sum=0;
    for(int i=0; i<len; i++)
    {
        if(join(e[i].u,e[i].v))
        {
            cont++;
            sum+=e[i].w;
        }
        if(cont==n-1)
            break;
    }
    printf("%.2f\n",sum);
}

int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        init();
        double w;
        len=0;
        for(int i=1; i<=n; i++)
        {
            scanf("%d%d",&x[i],&y[i]);
            for(int j=1; j<i; j++)
            {
                w=sqrt(1.0*(x[i]-x[j])*(x[i]-x[j])+1.0*(y[i]-y[j])*(y[i]-y[j]));
                e[len].u=i,e[len].v=j,e[len++].w=w;
//                e[len].u=j,e[len].v=i,e[len++].w=w;
            }
        }
        int u,v;
        for(int i=1; i<=m; i++)
        {
            scanf("%d%d",&u,&v);
            int fx=Find(u),fy=Find(v);
            pre[fx]=fy;
        }
        Kruskal();
    }
    return 0;
}

ps：建双向边c++超时，g++AC，建单向边都AC，（其实只建单向边就好了，因为Kruskal只是按边的长度来排序的，不管你是从那个点到那个点）

代码（Prim算法）：

#include<stdio.h>
#include<math.h>
#include<string.h>

#define maxn 1000+10
#define min(a,b) (a<b?a:b)
const double inf=0x3f3f3f3f;
double e[maxn][maxn],d[maxn];
bool vis[maxn];
int x[maxn],y[maxn];
int n,m;

void Prim()
{
    int i,j,v;
    for(int i=1; i<=n; i++)
        d[i]=e[1][i];
    vis[1]=1;
    double sum=0,minn;
    while(1)
    {
        minn=inf;
        for(i=1; i<=n; i++)
            if(!vis[i]&&d[i]<minn)
                minn=d[i],v=i;
        if(minn==inf)
            break;
        vis[v]=true,sum+=minn;
        for(i=1; i<=n; i++)
            if(!vis[i]&&d[i]>e[v][i])
                d[i]=e[v][i];
    }
    printf("%.2lf\n",sum);
}

int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=n; j++)
                e[i][j]=e[j][i]=(i==j?0:inf);
            d[i]=inf,vis[i]=0;
        }
        double w;
        for(int i=1; i<=n; i++)
        {
            scanf("%d%d",&x[i],&y[i]);
            for(int j=1; j<i; j++)
            {
                w=sqrt(1.0*(x[i]-x[j])*(x[i]-x[j])+1.0*(y[i]-y[j])*(y[i]-y[j]));
                e[i][j]=e[j][i]=w;
            }
        }
        int u,v;
        for(int i=1; i<=m; i++)
        {
            scanf("%d%d",&u,&v);
            e[u][v]=e[v][u]=0;
        }
        Prim();
    }
    return 0;
}

ps：做的时候忘记Prim算法的本质了，结果初始的时候我给建了一片森林，真是wrong的无厘头0.0，初始的时候只需要随便选一个点，接着逐步建立起一个最小生成树就好了