BZOJ1767: [Ceoi2009]harbingers

题目：http://www.lydsy.com/JudgeOnline/problem.php?id=1767

题解：果然NOI2014购票出了原题233 虽然加上距离限制之后麻烦了好多。。。

不过没有限制的话，直接把整个x-rt的凸包建出来，然后每个点都去二分即可。

代码：

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<string>
#define inf 1000000000
#define maxn 200000+5
#define maxm 100000+5
#define eps 1e-10
#define ll long long
#define pa pair<int,int>
#define for0(i,n) for(int i=0;i<=(n);i++)
#define for1(i,n) for(int i=1;i<=(n);i++)
#define for2(i,x,y) for(int i=(x);i<=(y);i++)
#define for3(i,x,y) for(int i=(x);i>=(y);i--)
#define for4(i,x) for(int i=head[x],y=e[i].go;i;i=e[i].next,y=e[i].go)if(e[i].can&&y!=fa[x])
#define mod 1000000007
using namespace std;
inline 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=10*x+ch-'0';ch=getchar();}
return x*f;
}
int n,cnt,top,sta[maxn],head[maxn],g[maxn],tot,root,s[maxn],sum,fa[maxn],ss[maxn];
ll d[maxn],p[maxn],q[maxn],f[maxn],lim[maxn];
double k[maxn];
struct edge{int go,next;ll w;bool can;}e[maxn];
inline void add(int x,int y,ll w)
{
e[++tot]=(edge){y,head[x],w,1};head[x]=tot;
e[++tot]=(edge){x,head[y],w,1};head[y]=tot;
}
inline void dfs(int x)
{
for4(i,x)d[y]=d[fa[y]=x]+e[i].w,dfs(y);
}
inline void getrt(int x)
{
ss[x]=0;s[x]=1;
for4(i,x){getrt(y);s[x]+=s[y];ss[x]=max(ss[x],s[y]);}
ss[x]=max(ss[x],sum-s[x]);
if(ss[x]<ss[root])root=x;
}
inline void get(int x)
{
g[++g[0]]=x;
for4(i,x)get(y);
}
inline double slope(int x,int y)
{
return (double)(f[x]-f[y])/(double)(d[x]-d[y]);
}
inline void  insert(int x)
{
while(top>1&&slope(x,sta[top])>slope(sta[top],sta[top-1]))top--;
sta[++top]=x;k[top]=-slope(x,sta[top-1]);
}
inline bool cmp(int x,int y){return d[x]-lim[x]>d[y]-lim[y];}
inline void use(int x,int y)
{
f[x]=min(f[x],f[y]+(d[x]-d[y])*p[x]+q[x]);
}
void solve(int x)
{
if(sum<=1)return;
root=0;getrt(x);int rt=root;
for4(i,fa[rt])if(y==rt){e[i].can=0;sum=s[x]-s[y];solve(x);break;}
for(int i=fa[rt];i!=fa[x];i=fa[i])use(rt,i);
g[0]=0;
for4(i,rt)get(y);
top=0;
for(int i=rt;i!=fa[x];i=fa[i])insert(i);
for1(i,g[0])use(g[i],sta[min(top,upper_bound(k+2,k+top+1,-p[g[i]])-k-1)]);
for4(i,rt){e[i].can=0;sum=s[y];solve(y);}
}
int main()
{
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
n=read();ss[0]=inf;
for1(i,n-1){int x=read(),y=read(),z=read();add(x,y,z);}
for2(i,2,n)q[i]=read(),p[i]=read(),f[i]=1ll<<62;
dfs(1);
sum=n;solve(1);
for2(i,2,n)printf("%lld%c",f[i],i==n?'\n':' ');
return 0;
}

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