Codeforces 161D Distance in Tree(点分治/树形dp)
题意:给出一个树,求有多少对节点间的距离等于k,n<5e4, k<500
题解:解法一:点分治,k很小,在拿一个桶统计一下,注意统计的时候要判断时候和本身相同,相同要减一
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <bits/stdc++.h> #define IO_read ios::sync_with_stdio(false);cin.tie(0) #define fre freopen("in.txt", "r", stdin) #define _for(i,a,b) for(int i=a; i< b; i++) #define _rep(i,a,b) for(int i=a; i<=b; i++) #define inf 0x3f3f3f3f #define lowbit(a) ((a)&-(a)) using namespace std; typedef long long ll; template <class T> void read(T &x) { char c; bool op=0; while(c=getchar(), c<'0'||c>'9') if(c=='-') op=1; x=c-'0'; while(c=getchar(), c>='0'&&c<='9') x=x*10+c-'0'; if(op) x=-x; } template <class T> void write(T x) { if(x<0) putchar('-'), x=-x; if(x>=10) write(x/10); putchar('0'+x%10); } const int maxn=5e4+5; int head[maxn], tot; struct Edge{ int to, next, val; }edge[maxn*2]; void addedge(int u, int v, int w) { edge[++tot]={v, head[u], w}; head[u]=tot; } int n, k; ll ans; int sz[maxn], vis[maxn], bucket[505], max_son, rt, Size; ll dis[maxn], q[maxn]; int l, r; void get_rt(int u, int fa) { sz[u]=1; int num=0; for(int i=head[u]; i!=-1; i=edge[i].next) { int v=edge[i].to; if(vis[v] || v==fa) continue; get_rt(v, u); sz[u]+=sz[v]; num=max(num, sz[v]); } num=max(num, Size-sz[u]); if(num<max_son) max_son=num, rt=u; } void get_dis(int u, int fa) { q[++r]=dis[u]; for(int i=head[u]; i!=-1; i=edge[i].next) { int v=edge[i].to, w=edge[i].val; if(vis[v] || v==fa) continue; dis[v]=dis[u]+w; get_dis(v, u); } } ll solve(int u, int val) { l=1, r=0, dis[u]=val; get_dis(u, 0); ll res=0; sort(q+1, q+1+r); memset(bucket, 0, sizeof(bucket)); for(int i=1; i<=r&&q[i]<=k; i++) bucket[q[i]]++; for(int i=1; i<=r&&q[i]<=k; i++) { if(k-q[i]==q[i]) res+=bucket[k-q[i]]-1; else res+=bucket[k-q[i]]; } /* for(int i=1; i<=r; i++) printf("%d ", q[i]); printf("\n"); printf("res=%d, k=%d\n", res); */ res=res/2; return res; } void divide(int u) { ans+=solve(u, 0); vis[u]=true; for(int i=head[u]; i!=-1; i=edge[i].next) { int v=edge[i].to, w=edge[i].val; if(vis[v]) continue; ans-=solve(v, w); //容斥 Size=sz[v], max_son=inf; get_rt(v, 0); divide(rt); } } int main() { IO_read; //fre; cin>>n>>k; memset(head, -1, sizeof(head)); for(int i=1,u,v; i<n; i++) { cin>>u>>v; addedge(u, v, 1), addedge(v, u, 1); } Size=n, max_son=inf; get_rt(1, 0); divide(rt); cout<<ans<<"\n"; }
解法二:树形dp,待解决