HDU 1875 畅通工程再续(MST)

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题目链接:
HDU 1875 畅通工程再续
分析:
把不符合条件的边忽略再检查下所有点都连通就行了。

//1676K 62MS#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;const int maxn=110;const int INF=0x3f3f3f3f;int T,n,tot;int pre[maxn];double ans;struct Point{    int x,y;}point[maxn];struct Edge{    int u,v;    double w;}edge[maxn*maxn];void init(){    for(int i=0;i<maxn;i++)        pre[i]=i;}int find(int x){    return pre[x]==x?x:pre[x]=find(pre[x]);}double Distance(struct Point a,struct Point b){    int xx=a.x-b.x;    int yy=a.y-b.y;    return sqrt(xx*xx+yy*yy);}bool cmp(struct Edge a,struct Edge b){    return a.w<b.w;}int main(){#ifdef LOCAL    freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);#endif    scanf("%d",&T);    while(T--)    {        init();        scanf("%d",&n);        for(int i=1;i<=n;i++)            scanf("%d%d",&point[i].x,&point[i].y);        tot=0;        for(int i=1;i<=n;i++)        {            for(int j=i+1;j<=n;j++)            {                edge[tot].u=i;                edge[tot].v=j;                edge[tot].w=Distance(point[i],point[j]);                tot++;            }        }        sort(edge,edge+tot,cmp);        ans=0;        for(int i=0;i<tot;i++)        {            if(edge[i].w>1000||edge[i].w<10) continue;            int u=edge[i].u;            int v=edge[i].v;            int fu=find(u);            int fv=find(v);            if(fu==fv) continue;            pre[fu]=fv;            ans+=edge[i].w;        }        int com=find(1);        int flag=0;        for(int i=2;i<=n;i++)        {            if(find(i)!=com)            {                flag=1;                break;            }        }        if(flag) printf("oh!\n");        else printf("%.1f\n",ans*100);    }    return 0;}
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