hdu 畅通工程再续（最小生成树）（Prim算法 && Kruskal算法）

2210 1020 2031 12 21000 1000 Sample Output
1414.2oh!代码（Kruskal算法）：
#include<stdio.h>#include<math.h>#include<string.h>#include<algorithm>using namespace std;#define maxn 100+10#define maxv 10000+10struct edge{    int u,v;    double w;} e[maxv];int pre[maxn],x[maxn],y[maxn];int n,len;bool cmp(edge a,edge b){    return a.w<b.w;}void init(){    for(int i=1; i<maxn; i++)        pre[i]=i;}int Find(int x){    if(x==pre[x])        return x;    else    {        pre[x]=Find(pre[x]);        return pre[x];    }}int join(int x,int y){    int fx=Find(x),fy=Find(y);    if(fx!=fy)    {        pre[fy]=fx;        return 1;    }    return 0;}void Kruskal(){    sort(e,e+len,cmp);    int cont=0;    double sum=0;    for(int i=1; i<len; i++)    {        if(e[i].w<10||e[i].w>1000)            continue;        if(join(e[i].u,e[i].v))        {            cont++;            sum+=e[i].w*100;        }        if(cont==n-1)            break;    }    if(cont<n-1)        printf("oh!/n");    else        printf("%.1lf/n",sum);}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        init();        int i,j;        double w;        len=1;        for(i=1; i<=n; i++)        {            scanf("%d%d",&x[i],&y[i]);            for(j=1; j<i; j++)            {                w=sqrt((double)(x[i]-x[j])*(double)(x[i]-x[j])+(double)(y[i]-y[j])*(double)(y[i]-y[j]));                e[len].u=i,e[len].v=j,e[len++].w=w;            }        }        Kruskal();    }    return 0;}代码（Prim算法）：
#include<stdio.h>#include<math.h>#include<string.h>#define maxn 100+10const double inf=0x3f3f3f3f;double e[maxn][maxn],d[maxn];bool vis[maxn];int x[maxn],y[maxn];int n;void Prim(){    int i,j,v;    double minn;    for(i=1; i<=n; i++)    {        d[i]=e[1][i];        vis[i]=0;    }    vis[1]=1;//已经找到了一个点    double sum=0;    for(j=1; j<n; j++)//还需找n-1个点    {        minn=inf;        for(i=1; i<=n; i++)            if(!vis[i]&&d[i]<minn)                minn=d[i],v=i;        if(minn==inf)        {            printf("oh!/n");            return ;        }        vis[v]=1,sum+=d[v];        for(i=1; i<=n; i++)            if(!vis[i]&&d[i]>e[v][i])                d[i]=e[v][i];    }    printf("%.1lf/n",sum*100);}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        int i,j;        double w;        for(i=1; i<=n; i++)            for(j=1; j<=n; j++)                i==j?e[i][j]=0:e[i][j]=e[j][i]=inf;        for(i=1; i<=n; i++)        {            scanf("%d%d",&x[i],&y[i]);            for(j=1; j<i; j++)            {                w=sqrt((double)(x[i]-x[j])*(double)(x[i]-x[j])+(double)(y[i]-y[j])*(double)(y[i]-y[j]));                if(w>1000||w<10)                    e[i][j]=e[j][i]=inf;                else                    e[i][j]=e[j][i]=w;            }        }        Prim();    }    return 0;}