题目链接:https://www.luogu.com.cn/problem/P2212
几乎是Kruskal裸题,但是建n*(n-1)条边给我T了俩点,后来我发现只要C(n,2)条边就可以,因为假设(vi,vj)和(vj,vi)之间有边,但是其中一条没用到则另外一条也用不到,因为他们一样长,如果其中一条边在生成树里面,另一条边也是不会用到的,因为不能有环。所以两点之间一条边就可以。这样时间就降了一半。另外,防T?快读?register?inline?orz orz
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef unsigned int ui; 4 typedef long long ll; 5 typedef unsigned long long ull; 6 #define pf printf 7 #define mem(a,b) memset(a,b,sizeof(a)) 8 #define prime1 1e9+7 9 #define prime2 1e9+9 10 #define pi 3.14159265 11 #define lson l,mid,rt<<1 12 #define rson mid+1,r,rt<<1|1 13 #define scand(x) scanf("%llf",&x) 14 #define f(i,a,b) for(int i=a;i<=b;i++) 15 #define scan(a) scanf("%d",&a) 16 #define mp(a,b) make_pair((a),(b)) 17 #define P pair<int,int> 18 #define dbg(args) cout<<#args<<":"<<args<<endl; 19 #define inf 0x3f3f3f3f 20 const int maxn=1e6+10; 21 int n,m,t; 22 inline int read(){ 23 int ans=0,w=1; 24 char ch=getchar(); 25 while(!isdigit(ch)){if(ch=='-')w=-1;ch=getchar();} 26 while(isdigit(ch))ans=(ans<<3)+(ans<<1)+ch-'0',ch=getchar(); 27 return ans*w; 28 } 29 struct point{ 30 int x,y; 31 }p[maxn]; 32 struct node{ 33 int u,v,w; 34 }e[maxn*5];//注意边的数量,因为越界wa了一发 35 int cnt=0; 36 int dis(int i,int j){return (p[i].x-p[j].x)*(p[i].x-p[j].x)+(p[i].y-p[j].y)*(p[i].y-p[j].y);} 37 bool cmp(node& a,node& b) 38 { 39 return a.w<b.w; 40 } 41 int f[maxn]; 42 int ans,tot; 43 void init() 44 { 45 f(i,1,n)f[i]=i; 46 tot=0; 47 ans=0; 48 } 49 int find(int x) 50 { 51 return x==f[x]?x:f[x]=find(f[x]); 52 } 53 void Union(int x,int y,int w) 54 { 55 int fx=find(x); 56 int fy=find(y); 57 if(fx==fy)return ; 58 else 59 { 60 if(w<m)return; 61 if(tot==n-1)return ; 62 f[fx]=fy; 63 ans+=w; 64 tot++; 65 } 66 } 67 int main() 68 { 69 //freopen("input.txt","r",stdin); 70 //freopen("output.txt","w",stdout); 71 std::ios::sync_with_stdio(false); 72 n=read(),m=read(); 73 init(); 74 f(i,1,n) 75 { 76 p[i].x=read(),p[i].y=read(); 77 } 78 f(i,1,n) 79 f(j,i+1,n) 80 { 81 if(i==j)continue; 82 e[++cnt].u=i; 83 e[cnt].v=j; 84 e[cnt].w=dis(i,j); 85 } 86 sort(e+1,e+cnt+1,cmp); 87 f(i,1,cnt) 88 { 89 if(e[i].w<m)continue; 90 Union(e[i].u,e[i].v,e[i].w); 91 } 92 if(tot!=n-1) 93 { 94 pf("-1"); 95 } 96 else 97 { 98 pf("%d",ans); 99 } 100 }