XJOI网上同步训练DAY3 T2

Posted 2020-07-31 GFY

考试的时候已经想出来怎么做了，但是没有时间打了T_T

思路:我们考虑将询问以lim排序，然后树链剖分，把边作为线段树的节点，然后随着询问lim的增大，改变线段树中节点的信息，然后每次询问我们用树链剖分询问，复杂度是O(nlogn）,又get一种新的树链剖分打法

  1 #include<cstdio>
  2 #include<cmath>
  3 #include<iostream>
  4 #include<algorithm>
  5 #include<cstring>
  6 struct node{
  7     int x,y,lim,id;
  8 }q[200005];
  9 int V[200005],dep[200005],t[200005],dfn[200005],num,n,m;
 10 struct Data{
 11     int l,r,s,v;
 12     Data(){}
 13     Data(int a,int b,int c,int d):l(a),r(b),s(c),v(d){}
 14 }s1[800005],s2[800005];
 15 Data operator +(Data a,Data b){
 16         Data ret(a.l,b.r,a.s+b.s,a.v+b.v);
 17         if (a.r&&b.l) ret.v=a.v+b.v-V[a.r]-V[b.l]+V[a.r+b.l];
 18         if (b.l==b.s) ret.r=a.r+b.s;
 19         if (a.l==a.s) ret.l=b.l+a.s;
 20         return ret;
 21 }
 22 int tot,go[200005],next[200005],first[200005];
 23 int size[200005],son[200005],top[200005],fa[200005];
 24 int ans[200005],val[200005];
 25 int read(){
 26     int t=0,f=1;char ch=getchar();
 27     while (ch<\'0\'||ch>\'9\') {if (ch==\'-\') f=-1;ch=getchar();}
 28     while (\'0\'<=ch&&ch<=\'9\'){t=t*10+ch-\'0\';ch=getchar();}
 29     return t*f;
 30 }
 31 void insert(int x,int y,int z){
 32     tot++;
 33     go[tot]=y;
 34     next[tot]=first[x];
 35     first[x]=tot;
 36     val[tot]=z;
 37 }
 38 void add(int x,int y,int z){
 39     insert(x,y,z);insert(y,x,z);
 40 }
 41 bool cmp(node a,node b){
 42     return a.lim<b.lim;
 43 }
 44 bool cmp1(int x,int y){
 45     return val[x]<val[y];
 46 }
 47 void build(int k,int l,int r){
 48     if (l==r){
 49         s1[k]=s2[k]=Data(1,1,1,V[1]);
 50         return;
 51     }
 52     int mid=(l+r)/2;
 53     build(k*2,l,mid);
 54     build(k*2+1,mid+1,r);
 55     s1[k]=s1[k*2]+s1[k*2+1];
 56     s2[k]=s2[k*2+1]+s2[k*2];
 57 }
 58 void dfs1(int x,int f){
 59     size[x]=1;
 60     for (int i=first[x];i;i=next[i]){
 61         int pur=go[i];
 62         if (pur!=f){
 63             dep[pur]=dep[x]+1;
 64             t[++t[0]]=i;
 65             dfs1(pur,x);
 66             size[x]+=size[pur];
 67             if (size[pur]>size[son[x]]) son[x]=pur;
 68         }
 69     }
 70 }
 71 void dfs2(int x,int f){
 72     dfn[x]=++num;
 73     if (son[x]) top[son[x]]=top[x],dfs2(son[x],x);
 74     for (int i=first[x];i;i=next[i]){
 75         int pur=go[i];
 76         if (pur==f||pur==son[x]) continue;
 77         top[pur]=pur;
 78         fa[pur]=x;
 79         dfs2(pur,x);
 80     }
 81 }
 82 void modify(int k,int l,int r,int pos){
 83     if (l==r){
 84         s1[k]=s2[k]=Data(0,0,1,0);
 85         return;
 86     }
 87     int mid=(l+r)/2;
 88     if (pos<=mid) modify(k*2,l,mid,pos);
 89     else modify(k*2+1,mid+1,r,pos);
 90     s1[k]=s1[k*2]+s1[k*2+1];
 91     s2[k]=s2[k*2+1]+s2[k*2];
 92 }
 93 Data ask1(int k,int l,int r,int x,int y){
 94     if (l==x&&r==y){
 95         return s1[k];
 96     }
 97     int mid=(l+r)/2;
 98     if (y<=mid) return ask1(k*2,l,mid,x,y);
 99     else
100     if (x>mid) return ask1(k*2+1,mid+1,r,x,y);
101     else return ask1(k*2,l,mid,x,mid)+ask1(k*2+1,mid+1,r,mid+1,y);
102 }
103 Data ask2(int k,int l,int r,int x,int y){
104     if (l==x&&r==y){
105         return s2[k];
106     }
107     int mid=(l+r)/2;
108     if (y<=mid) return ask2(k*2,l,mid,x,y);
109     else
110     if (x>mid) return ask2(k*2+1,mid+1,r,x,y);
111     else return ask2(k*2+1,mid+1,r,mid+1,y)+ask2(k*2,l,mid,x,mid);
112 }
113 int work(int x,int y){
114     Data ans1(0,0,0,0),ans2(0,0,0,0);
115     while (top[x]!=top[y]){
116         if (dep[top[x]]<dep[top[y]]){
117             ans2=ask1(1,1,n,dfn[top[y]],dfn[y])+ans2;
118             y=fa[top[y]];
119         }else{
120             ans1=ans1+ask2(1,1,n,dfn[top[x]],dfn[x]);
121             x=fa[top[x]];
122         }
123     }
124     if (x!=y){
125         if (dep[x]<dep[y]){
126             ans2=ask1(1,1,n,dfn[son[x]],dfn[y])+ans2;
127         }else{
128             ans1=ans1+ask2(1,1,n,dfn[son[y]],dfn[x]);
129         }
130     }
131     return (ans1+ans2).v;
132 }
133 int main(){
134     freopen("tx.in","r",stdin);
135     n=read();
136     for (int i=1;i<n;i++) V[i]=read();
137     for (int i=1;i<n;i++){
138         int x=read()+1,y=read()+1,v=read();
139         add(x,y,v);
140     }
141     m=read();
142     for (int i=1;i<=m;i++){
143         q[i].x=read()+1,q[i].y=read()+1,q[i].lim=read();q[i].id=i;
144     }
145     dfs1(1,0);dfs2(1,0);
146     std::sort(q+1,q+1+m,cmp);
147     std::sort(t+1,t+1+t[0],cmp1);
148     int h=1;build(1,1,n);
149     for (int i=1;i<=m;i++){
150         while (val[t[h]]<=q[i].lim&&h<=t[0]){
151             modify(1,1,n,dfn[go[t[h]]]);
152             h++;
153         }
154         ans[q[i].id]=work(q[i].x,q[i].y);
155     }
156     for (int i=1;i<=m;i++) printf("%d\\n",ans[i]);
157 }