题意
分析
考场做法
一眼看出是支持换根的树形dp。
用(f(x,0/1))表示x及其子树中，从x出发，不一定/一定回到x的最大收益。
然后子树很好做。
换根的时候，我先计算后还原，需要考虑很多，调了很久。
后来知道可以用up,down状态转移，会好写一些，但要考虑得跟我先前打的差不多。
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<iostream>
#include<string>
#include<vector>
#include<list>
#include<deque>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<bitset>
#include<algorithm>
#include<complex>
#include<cassert>
#define rg register
#define il inline
#define co const
#pragma GCC optimize ("O0")
using namespace std;
template<class T> il T read()
{
T data=0;
int w=1;
char ch=getchar();
while(!isdigit(ch))
{
if(ch=='-')
w=-1;
ch=getchar();
}
while(isdigit(ch))
data=10*data+ch-'0',ch=getchar();
return data*w;
}
template<class T> il T read(T&x)
{
return x=read<T>();
}
typedef long long ll;
const int INF=0x7fffffff;
co int MAXN=3e5+7;
struct Edge
{
int nx,to,w;
}E[MAXN<<1];
int head[MAXN],ecnt;
il void addedge(rg int x,rg int y,rg int w)
{
E[++ecnt].to=y,E[ecnt].w=w;
E[ecnt].nx=head[x],head[x]=ecnt;
}
int fa[MAXN],fw[MAXN];
int val[MAXN];
ll f[MAXN][2];
int maxv[MAXN],secv[MAXN];
il ll take(rg int y)
{
return max(0LL,f[y][0]-fw[y]);
}
il ll fetch(rg int y)
{
return max(0LL,f[y][1]-2*fw[y]);
}
il ll calc(rg int y)
{
return take(y)-fetch(y);
}
il void dfs1(rg int x,rg int fath)
{
fa[x]=fath;
f[x][1]=val[x];
for(rg int i=head[x];i;i=E[i].nx)
{
rg int y=E[i].to,w=E[i].w;
if(y==fath)
{
fw[x]=w;
continue;
}
dfs1(y,x);
f[x][1]+=fetch(y);
}
f[x][0]=f[x][1];
for(rg int i=head[x];i;i=E[i].nx)
{
rg int y=E[i].to;
if(y==fath)
continue;
if(calc(y)>calc(secv[x]))
{
secv[x]=y;
if(calc(secv[x])>calc(maxv[x]))
swap(secv[x],maxv[x]);
}
}
f[x][0]+=calc(maxv[x]);
}
ll fet[MAXN],calsec[MAXN],cal[MAXN];
il void dfs2(rg int x)
{
if(fa[x])
{ // f[fa] shouldn't change
fet[x]=fetch(x);
cal[x]=calc(x);
f[fa[x]][1]-=fet[x];
f[fa[x]][0]-=fet[x];
rg int w=fw[fa[x]];
fw[fa[x]]=fw[x];
f[x][1]+=fetch(fa[x]);
f[x][0]=f[x][1];
if(maxv[fa[x]]==x)
{
f[fa[x]][0]-=cal[x];
f[fa[x]][0]+=calsec[fa[x]];
}
if(calc(fa[x])>calc(secv[x]))
{
secv[x]=fa[x];
if(calc(secv[x])>calc(maxv[x]))
swap(secv[x],maxv[x]);
}
calsec[x]=calc(secv[x]);
f[x][0]+=calc(maxv[x]);
f[fa[x]][1]+=fet[x]; // restore
f[fa[x]][0]+=fet[x];
fw[fa[x]]=w;
if(maxv[fa[x]]==x)
{
f[fa[x]][0]-=calsec[fa[x]];
f[fa[x]][0]+=cal[x];
}
}
else
{
fet[x]=fetch(x);
calsec[x]=calc(secv[x]);
cal[x]=calc(x);
}
for(rg int i=head[x];i;i=E[i].nx)
{
rg int y=E[i].to;
if(y==fa[x])
continue;
dfs2(y);
}
}
int main()
{
freopen("treasure.in","r",stdin);
freopen("treasure.out","w",stdout);
rg int n=read<int>();
for(rg int i=1;i<n;++i)
{
rg int x=read<int>(),y=read<int>(),w=read<int>();
addedge(x,y,w);
addedge(y,x,w);
}
for(rg int i=1;i<=n;++i)
{
read(val[i]);
}
dfs1(1,0);
dfs2(1);
/* for(int i=1;i<=n;++i)
{
printf("%d:
",i);
printf(" f0=%lld f1=%lld
",f[i][0],f[i][1]);
}*/
for(rg int i=1;i<=n;++i)
{
printf("%lld
",f[i][0]);
}
// fclose(stdin);
// fclose(stdout);
return 0;
}
标解
#include <iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<bitset>
using namespace std;
typedef long long ll;
const int N=610000;
int fi[N],ne[N],len[N],to[N];
ll dp0[N],dp1[N];
ll ans[N],wushi0[N],wushi1[N],val[N];
int pE,n;
//0：考虑回到原点；1：不回到原点
void addE(int u,int v,int l)
{
pE++;to[pE]=v;len[pE]=l;
ne[pE]=fi[u];fi[u]=pE;
}
void dfs(int u,int fa)
{
ll mx=0;
for (int j=fi[u];j;j=ne[j])
{
if (to[j]==fa) continue;
dfs(to[j],u);
if (dp0[to[j]]-2*len[j]>0)
{
dp0[u]+=dp0[to[j]]-2*len[j];
mx=max(mx,len[j]+dp1[to[j]]-dp0[to[j]]);
}
else mx=max(mx,dp1[to[j]]-len[j]);
}
dp0[u]+=val[u];
dp1[u]=dp0[u]+mx;
}
void dfs2(int u,int fa,int le)
{
ans[u]=max(dp0[u]+max(wushi1[u]-le,(ll)0),dp1[u]+max(wushi0[u]-le*2,(ll)0));
ll mx1=0,mx2=0;
for (int j=fi[u];j;j=ne[j])
{
if (to[j]==fa) continue;
ll now;
if (dp0[to[j]]-2*len[j]>0) now=len[j]+dp1[to[j]]-dp0[to[j]];
else now=dp1[to[j]]-len[j];
if (now>mx1) {mx2=mx1;mx1=now;}
else if (now>mx2) mx2=now;
}
ll now;
if (wushi0[u]-2*le>0) now=le+wushi1[u]-wushi0[u];
else now=wushi1[u]-le;
if (now>mx1) {mx2=mx1;mx1=now;}
else if (now>mx2) mx2=now;
for (int j=fi[u];j;j=ne[j])
{
if (to[j]==fa) continue;
wushi0[to[j]]=dp0[u]-max((ll)0,dp0[to[j]]-2*len[j])+max((ll)0,wushi0[u]-2*le);
ll now;
if (dp0[to[j]]-2*len[j]>0) now=len[j]+dp1[to[j]]-dp0[to[j]];
else now=dp1[to[j]]-len[j];
if (now==mx1) wushi1[to[j]]=wushi0[to[j]]+mx2;
else wushi1[to[j]]=wushi0[to[j]]+mx1;
}
for (int j=fi[u];j;j=ne[j])
if (to[j]!=fa) dfs2(to[j],u,len[j]);
}
int main()
{
freopen("treasure.in","r",stdin);
freopen("treasure.out","w",stdout);
scanf("%d",&n);
for (int i=1;i<n;i++)
{
int u,v,l;scanf("%d%d%d",&u,&v,&l);
addE(u,v,l);addE(v,u,l);
}
for (int i=1;i<=n;i++) scanf("%d",&val[i]);
dfs(1,0);
for (int j=fi[1];j;j=ne[j])
if (dp0[to[j]]-2*len[j]>0)
wushi0[to[j]]=dp0[1]-(dp0[to[j]]-2*len[j]);
else wushi0[to[j]]=dp0[1];
ll mx1=0,mx2=0;
for (int j=fi[1];j;j=ne[j])
{
ll now;
if (dp0[to[j]]-2*len[j]>0) now=len[j]+dp1[to[j]]-dp0[to[j]];
else now=dp1[to[j]]-len[j];
if (now>mx1) {mx2=mx1;mx1=now;}
else if (now>mx2) mx2=now;
}
for (int j=fi[1];j;j=ne[j])
{
ll now;
if (dp0[to[j]]-2*len[j]>0) now=len[j]+dp1[to[j]]-dp0[to[j]];
else now=dp1[to[j]]-len[j];
if (now==mx1) wushi1[to[j]]=wushi0[to[j]]+mx2;
else wushi1[to[j]]=wushi0[to[j]]+mx1;
}
ans[1]=dp1[1];
for (int j=fi[1];j;j=ne[j]) dfs2(to[j],1,len[j]);
for (int i=1;i<=n;i++) printf("%lld
",ans[i]);
return 0;
}