点分治裸题23333这个题如果K很大的话就不能用依次处理每棵子树做了。（数组存不开2333）
这个就只能是总的依次减去每棵子树里的。
1 #include<cstdio> 2 #include<algorithm> 3 #include<iostream> 4 #include<cstdlib> 5 #include<cstring> 6 #include<cmath> 7 #define N 100005 8 #define M 1000005 9 #define LL long long 10 #define inf 0x3f3f3f3f 11 using namespace std; 12 inline int ra() 13 { 14 int x=0,f=1; char ch=getchar(); 15 while (ch<'0' || ch>'9') {if (ch=='-') f=-1; ch=getchar();} 16 while (ch>='0' && ch<='9') {x=x*10+ch-'0'; ch=getchar();} 17 return x*f; 18 } 19 int n,K,cnt,sum,ans,root; 20 int head[10005],deep[10005],d[10005],f[10005],son[10005]; 21 bool vis[10005]; 22 struct data{int to,next,v;}e[20005]; 23 void insert(int x, int y, int w) 24 { 25 e[++cnt].next=head[x]; e[cnt].to=y; e[cnt].v=w; head[x]=cnt; 26 } 27 void getroot(int x, int fa) 28 { 29 son[x]=1; f[x]=0; 30 for (int i=head[x];i;i=e[i].next) 31 { 32 if (e[i].to==fa || vis[e[i].to]) continue; 33 getroot(e[i].to,x); 34 son[x]+=son[e[i].to]; 35 f[x]=max(f[x],son[e[i].to]); 36 } 37 f[x]=max(f[x],sum-son[x]); 38 if (f[x]<f[root]) root=x; 39 } 40 void getdeep(int x, int fa) 41 { 42 deep[++deep[0]]=d[x]; 43 for (int i=head[x];i;i=e[i].next) 44 { 45 if (e[i].to==fa || vis[e[i].to]) continue; 46 d[e[i].to]=d[x]+e[i].v; 47 getdeep(e[i].to,x); 48 } 49 } 50 int cal(int x, int now) 51 { 52 d[x]=now; deep[0]=0; 53 getdeep(x,0); 54 sort(deep+1,deep+deep[0]+1); 55 int t=0,l,r; 56 for (l=1,r=deep[0];l<r;) 57 { 58 if (deep[l]+deep[r]<=K) t+=r-l,l++; 59 else r--; 60 } 61 return t; 62 } 63 void work(int x) 64 { 65 ans+=cal(x,0); vis[x]=1; 66 for (int i=head[x];i;i=e[i].next) 67 { 68 if (vis[e[i].to]) continue; 69 ans-=cal(e[i].to,e[i].v); 70 sum=son[e[i].to]; 71 root=0; 72 getroot(e[i].to,root); 73 work(root); 74 } 75 } 76 int main() 77 { 78 while (1) 79 { 80 ans=0,root=0; cnt=0; 81 memset(vis,0,sizeof(vis)); 82 memset(head,0,sizeof(head)); 83 n=ra(); K=ra(); if (n==0) break; 84 for (int i=1; i<n; i++) 85 { 86 int x=ra(),y=ra(),w=ra(); 87 insert(x,y,w); insert(y,x,w); 88 } 89 sum=n; f[0]=inf; 90 getroot(1,0); 91 work(root); 92 cout<<ans<<endl; 93 } 94 return 0; 95 }