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：

 1 #include<cstdio>
 2 #include<cstdlib>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<algorithm>
 6 #include<iostream>
 7 #include<vector>
 8 #include<map>
 9 #include<set>
10 #include<queue>
11 #include<string>
12 #define inf 1000000000
13 #define maxn 200000+5
14 #define maxm 100000+5
15 #define eps 1e-10
16 #define ll long long
17 #define pa pair<int,int>
18 #define for0(i,n) for(int i=0;i<=(n);i++)
19 #define for1(i,n) for(int i=1;i<=(n);i++)
20 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
21 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
22 #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])
23 #define mod 1000000007
24 using namespace std;
25 inline ll read()
26 {
27     ll x=0,f=1;char ch=getchar();
28     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
29     while(ch>=‘0‘&&ch<=‘9‘){x=10*x+ch-‘0‘;ch=getchar();}
30     return x*f;
31 }
32 int n,cnt,top,sta[maxn],head[maxn],g[maxn],tot,root,s[maxn],sum,fa[maxn],ss[maxn];
33 ll d[maxn],p[maxn],q[maxn],f[maxn],lim[maxn];
34 double k[maxn];
35 struct edge{int go,next;ll w;bool can;}e[maxn];
36 inline void add(int x,int y,ll w)
37 {
38     e[++tot]=(edge){y,head[x],w,1};head[x]=tot;
39     e[++tot]=(edge){x,head[y],w,1};head[y]=tot;
40 }
41 inline void dfs(int x)
42 {
43     for4(i,x)d[y]=d[fa[y]=x]+e[i].w,dfs(y);
44 }
45 inline void getrt(int x)
46 {
47     ss[x]=0;s[x]=1;
48     for4(i,x){getrt(y);s[x]+=s[y];ss[x]=max(ss[x],s[y]);}
49     ss[x]=max(ss[x],sum-s[x]);
50     if(ss[x]<ss[root])root=x;
51 }
52 inline void get(int x)
53 {
54     g[++g[0]]=x;
55     for4(i,x)get(y);
56 }
57 inline double slope(int x,int y)
58 {
59     return (double)(f[x]-f[y])/(double)(d[x]-d[y]);
60 }
61 inline void  insert(int x)
62 {
63     while(top>1&&slope(x,sta[top])>slope(sta[top],sta[top-1]))top--;
64     sta[++top]=x;k[top]=-slope(x,sta[top-1]);
65 }
66 inline bool cmp(int x,int y){return d[x]-lim[x]>d[y]-lim[y];}
67 inline void use(int x,int y)
68 {
69     f[x]=min(f[x],f[y]+(d[x]-d[y])*p[x]+q[x]);
70 }
71 void solve(int x)
72 {
73     if(sum<=1)return;
74     root=0;getrt(x);int rt=root;
75     for4(i,fa[rt])if(y==rt){e[i].can=0;sum=s[x]-s[y];solve(x);break;}
76     for(int i=fa[rt];i!=fa[x];i=fa[i])use(rt,i);
77     g[0]=0;
78     for4(i,rt)get(y);
79     top=0;
80     for(int i=rt;i!=fa[x];i=fa[i])insert(i);
81     for1(i,g[0])use(g[i],sta[min(top,upper_bound(k+2,k+top+1,-p[g[i]])-k-1)]);
82     for4(i,rt){e[i].can=0;sum=s[y];solve(y);}
83 }
84 int main()
85 {
86     freopen("input.txt","r",stdin);
87     freopen("output.txt","w",stdout);
88     n=read();ss[0]=inf;
89     for1(i,n-1){int x=read(),y=read(),z=read();add(x,y,z);}
90     for2(i,2,n)q[i]=read(),p[i]=read(),f[i]=1ll<<62;
91     dfs(1);
92     sum=n;solve(1);
93     for2(i,2,n)printf("%lld%c",f[i],i==n?‘\n‘:‘ ‘);
94     return 0;
95 }