bzoj3479 [Usaco2014 Mar]Watering the Fields（prim裸题）

prim,O(n^2)求最小生成树。每次贪心的把最小边加进去。

#include 
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
#define pa pair
#define N 2010
inline int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*f;
}
int n,c,tot=0,d[N],ans=0;
struct node{int x,y;}a[N];
bool vis[N];
void prim(){
    priority_queuevector,greater >q;memset(d,0x3f,sizeof(d));
    d[1]=0;q.push(make_pair(0,1));
    while(!q.empty()){
        int x=q.top().second;q.pop();
        if(vis[x]) continue;vis[x]=1;++tot;ans+=d[x];
        for(int i=1;i<=n;++i){
            if(vis[i]) continue;int dx=(a[x].x-a[i].x)*(a[x].x-a[i].x)+(a[x].y-a[i].y)*(a[x].y-a[i].y);
            if(dx>=c&&dxint main(){
//  freopen("a.in","r",stdin);
    n=read();c=read();
    for(int i=1;i<=n;++i) a[i].x=read(),a[i].y=read();
    prim();
    if(tot!=n) puts("-1");
    else printf("%d\n",ans);
    return 0;
}