题目：

畅通工程再续

2210 1020 2031 12 21000 1000 Sample Output
1414.2oh! Author8600 Source2008浙大研究生复试热身赛（2）——全真模拟 Recommend思路：
典型的kruskal算法和prime算法，直接看注释
代码1（Kruskal）:
#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <string>#include <iostream>#include <stack>#include <queue>#include <vector>#include <algorithm>#define mem(a,b) memset(a,b,sizeof(a))#define N 100+20#define M 10000+20#define MOD 1000000000+7#define inf 0x3f3f3f3fusing namespace std;int n,len;int pre[N],x[N],y[N];struct node{    int u,v;    double w;} map[M];bool cmp(node a,node b){    return a.w<b.w;}void init(){    for(int i=1; i<=n; i++)        pre[i]=i;}int find(int x){    if(x==pre[x])        return x;    else    {        pre[x]=find(pre[x]);        return pre[x];    }}int mix(int x,int y){    int fx=find(x);    int fy=find(y);    if(fx!=fy)    {        pre[fy]=fx;        return 1;    }    return 0;}void Kruskal(){    sort(map,map+len,cmp);    int cnt=0;    double sum=0;    for(int i=1; i<=len; i++)    {        if(map[i].w<10||map[i].w>1000)            continue;        if(mix(map[i].u,map[i].v))//判断是否已经连接        {            cnt++;            sum+=map[i].w*100;        }        if(cnt==n-1)            break;    }    if(cnt<n-1)        printf("oh!/n");    else        printf("%.1lf/n",sum);}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        init();        double w;        len=1;        for(int i=1; i<=n; i++)        {            scanf("%d%d",&x[i],&y[i]);            for(int j=1; j<i; j++)//枚举所有的岛            {                w=sqrt((double)(x[i]-x[j])*(double)(x[i]-x[j])+(double)(y[i]-y[j])*(double)(y[i]-y[j]));                map[len].u=i;                map[len].v=j;                map[len].w=w;                len++;            }        }        Kruskal();    }    return 0;}代码2(Prime):
#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <string>#include <iostream>#include <stack>#include <queue>#include <vector>#include <algorithm>#define mem(a,b) memset(a,b,sizeof(a))#define N 100+20#define M 10000+20#define MOD 1000000000+7#define inf 0x3f3f3f3fusing namespace std;int n,x[N],y[N];bool vis[N];double map[N][N],dis[N];void Prim(){    double minn;    int k;    for(int i=1; i<=n; i++)    {        dis[i]=map[1][i];        vis[i]=0;    }    vis[1]=1;    double sum=0;    for(int i=1; i<=n-1; i++)//一共要进行n-1次    {        minn=inf;        for(int j=1; j<=n; j++)        {            if(!vis[j]&&dis[j]<minn)            {                minn=dis[j];                k=j;            }        }        if(minn==inf)//minn的值没有改变的话，那就是没有符合条件的        {            printf("oh!/n");            return;        }        vis[k]=1;        sum+=dis[k];//把选定的加上        for(int j=1; j<=n; j++)            if(!vis[j]&&map[k][j]<dis[j])                dis[j]=map[k][j];    }    printf("%.1lf/n",sum*100);}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        double w;        for(int i=1; i<=n; i++)            for(int j=1; j<=n; j++)                i==j?map[i][j]=0:map[i][j]=inf;        for(int i=1; i<=n; i++)        {            scanf("%d%d",&x[i],&y[i]);            for(int j=1; j<i; j++)//计算多组数据的距离            {                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)//不满足条件的直接设为无穷大                    map[i][j]=map[j][i]=inf;                else                    map[i][j]=map[j][i]=w;            }        }        Prim();    }    return 0;}