【bzoj1827】[Usaco2010 Mar]gather 奶牛大集会 树形dp+贪心

由x转移到y的答案为ans=ans+val∗(s[1]−s[y]∗2)
可以看出只有s[1]−s[y]∗2<0答案才会减小,所以贪心即可

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#include<vector>#define fo(i,a,b) for(int i=a;i<=b;i++)#define fod(i,a,b) for(int i=a;i>=b;i--)using namespace std;typedef long long ll;const int N=1e5+10;ll read(){    ll 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;}ll ans,C[N],dis[N],s[N];int n;vector<int>G[N],v[N];ll dfs(int x,int fa){    s[x]=C[x];    ll tot=dis[x]*C[x];    for(int i=0;i<G[x].size();i++)    {        int id=G[x][i],w=v[x][i];        if(id==fa)continue;        dis[id]=dis[x]+w;        tot+=dfs(id,x);        dfs(id,x);        s[x]+=s[id];    }    return tot;}void solve(int x,int fa){    for(int i=0;i<G[x].size();i++)    {        int id=G[x][i],w=v[x][i];        if(id==fa)continue;        if((s[id]<<1)>s[1]){ans=ans+(w*(s[1]-(s[id]<<1)));solve(id,x);}    }}int main(){    n=read();    fo(i,1,n)C[i]=read();    for(int from,to,val,i=1;i<n;i++)    {        from=read();to=read();val=read();        G[from].push_back(to);G[to].push_back(from);        v[from].push_back(val);v[to].push_back(val);    }     ans=dfs(1,0);       solve(1,0);    printf("%lld\n",ans);    return 0;}//PS:莫名T了

附上一开始没思路,打的O(n2logn)的树上倍增暴力
T_T比裸暴力还多一个log(┬＿┬)

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#define fo(i,a,b) for(int i=a;i<=b;i++)#define fod(i,a,b) for(int i=a;i>=b;i--)using namespace std;const int N=1e5+10,inf=(1<<30);int f[N][25],dis[N],last[N],dep[N],len=0;int n,C[N];struct Edge{int to,next,val;Edge(int to=0,int next=0,int val=0):to(to),next(next),val(val){}}e[N<<1];void add_edge(int u,int v,int w){e[++len]=Edge(v,last[u],w);last[u]=len;}void dfs(int x,int fa){    for(int i=1;i<=20;i++)     {        if(dep[x]<(1<<i))break;        f[x][i]=f[f[x][i-1]][i-1];    }    for(int i=last[x];i;i=e[i].next)     {        if(e[i].to==fa)continue;        f[e[i].to][0]=x;        dep[e[i].to]=dep[x]+1;        dis[e[i].to]=dis[x]+e[i].val;        dfs(e[i].to,x);    }}int lca(int x,int y){    if(dep[x]<dep[y])swap(x,y);    int d=dep[x]-dep[y];    for(int i=0;i<=20;i++)        if(d&(1<<i))x=f[x][i];    for(int i=20;i+1;i--)        if(f[x][i]!=f[y][i])            x=f[x][i],y=f[y][i];    if(x==y)return x;    else return f[x][0];}int main(){    scanf("%d",&n);    fo(i,1,n)scanf("%d",&C[i]);    for(int u,v,w,i=1;i<n;i++)     {        scanf("%d%d%d",&u,&v,&w);        add_edge(u,v,w);add_edge(v,u,w);    }    dfs(1,0);    int ans=inf;    for(int i=1;i<=n;i++)    {         int tmp=0;        for(int j=1;j<=n;j++)            tmp+=(dis[i]+dis[j]-2*dis[lca(i,j)])*C[j];        ans=min(ans,tmp);    }    printf("%d\n",ans);    return 0;}