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树分治。对于一棵子树的根节点,至少有一条边与儿子相连的属于重边。对于一条轻边,它的贡献值是两端子树大小的乘积,所以,重边应该是贡献值最大的一边。
至于要求所有的点,进行深度优先搜索,因为移动一个点只会影响两个点的两个子树,这个可以维护。
在进行DP时,选择计算最大的重边的值,答案就是用所有的边贡献值减去树的重边值的和。
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #define LL long long using namespace std; const int MAX=100010; int head[MAX],tol; struct Edge{ int u,v,next; }edge[MAX*2]; int n; void addedge(int u,int v){ edge[tol].u=u; edge[tol].v=v; edge[tol].next=head[u]; head[u]=tol++; } int counts[MAX]; LL dp[MAX]; int weight1[MAX],weight2[MAX]; LL ans[MAX],tot[MAX],par[MAX]; void dfs(int u,int f){ counts[u]=1; weight1[u]=weight2[u]=-1; for(int e=head[u];e!=-1;e=edge[e].next){ int v=edge[e].v; if(v!=f){ dfs(v,u); counts[u]+=counts[v]; dp[u]+=dp[v]; if(weight1[u]==-1||(LL)(n-counts[v])*counts[v]>(LL)(n-counts[weight1[u]])*counts[weight1[u]]){ weight2[u]=weight1[u]; weight1[u]=v; } else if(weight2[u]==-1||(LL)(n-counts[v])*counts[v]>(LL)(n-counts[weight2[u]])*counts[weight2[u]]){ weight2[u]=v; } /// dp[u]+=(n-counts[v])*counts[v]; } } tot[u]=dp[u]; if(counts[u]>1){ dp[u]+=(LL)(n-counts[weight1[u]])*counts[weight1[u]]; } } void dfs_ans(int u,int f){ for(int e=head[u];e!=-1;e=edge[e].next){ int v=edge[e].v; if(v==f) continue; if(v==weight1[u]){ LL num=max((LL)(n-counts[u])*counts[u],(LL)(n-counts[weight2[u]])*counts[weight2[u]]); par[v]=tot[u]-dp[v]+par[u]+num; } else{ LL num=max((LL)(n-counts[u])*counts[u],(LL)(n-counts[weight1[u]])*counts[weight1[u]]); par[v]=tot[u]-dp[v]+par[u]+num; } dfs_ans(v,u); } LL num=max((LL)(n-counts[weight1[u]])*counts[weight1[u]],(LL)(n-counts[u])*counts[u]); ans[u]=par[u]+tot[u]+num; } int main(){ while(scanf("%d",&n)!=EOF){ for(int i=1;i<=n;i++) head[i]=-1,counts[i]=0,dp[i]=0,par[i]=0; tol=0; int u,v; for(int i=1;i<n;i++){ scanf("%d%d",&u,&v); addedge(u,v); addedge(v,u); } dfs(1,0); dfs_ans(1,0); LL ans_tot=0; for(int i=1;i<=n;i++) ans_tot+=(LL)counts[i]*(n-counts[i]); for(int i=1;i<=n;i++) cout<<ans_tot-ans[i]<<endl; } return 0; }
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原文地址:http://www.cnblogs.com/jie-dcai/p/4957756.html