[bzoj 3052][wc2013]糖果公园

Posted 2020-09-20

传送门:http://www.lydsy.com/JudgeOnline/problem.php?id=3052

[wc2013]糖果公园

Time Limit: 200 Sec  Memory Limit: 512 MB
Submit: 1213  Solved: 609
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Description

 

Input

Output

Sample Input

Sample Input

 

Sample Output

84
131
27
84

树上莫队，把树分块，然后暴力，说起来是挺简单的，然而……

  1 #include<cstdio>
  2 #include<algorithm>
  3 #include<cmath>
  4 //using namespace std;
  5 const int maxn = 100005;
  6 typedef long long ll;
  7 ll ans,res[100005];
  8 int n,m,q;
  9 ll V[maxn],W[maxn],C[maxn];
 10 int h[maxn],nx[maxn<<1],to[maxn<<1],cnt;
 11 int fa[maxn],dep[maxn],sz[maxn],ch[maxn],top[maxn],num[maxn];
 12 int dfn[maxn],stack[maxn],stack_top,dfs_clock;
 13 int pos[maxn],pos_count,block_size,block_count,pre[maxn];
 14 bool vis[maxn];
 15 struct TAG1{
 16     int x,y,t,id;
 17 }b[maxn];
 18 struct TAG2{
 19     int x,y,t,pre;
 20 }c[maxn];
 21 int read(){
 22     int rt=0,fl=1;char ch=getchar();
 23     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)fl=-1;ch=getchar();}
 24     while(ch>=‘0‘&&ch<=‘9‘){rt=rt*10+ch-‘0‘;ch=getchar();}
 25     return rt*fl;
 26 }
 27 bool operator < (TAG1 a,TAG1 b){
 28     if(pos[a.x]==pos[b.x]&&pos[a.y]==pos[b.y])return a.t<b.t;
 29     else if(pos[a.x]==pos[b.x])return pos[a.y]<pos[b.y];
 30     else return pos[a.x]<pos[b.x];
 31 }
 32 void add_edge(int u,int v){
 33     to[++cnt]=v;nx[cnt]=h[u];h[u]=cnt;
 34     to[++cnt]=u;nx[cnt]=h[v];h[v]=cnt;
 35 }
 36 void dfs1(int x,int pr){
 37     fa[x]=pr;dep[x]=dep[pr]+1;sz[x]=1;
 38     for(int i=h[x];i;i=nx[i]){
 39         if(to[i]==pr)continue;
 40         dfs1(to[i],x);
 41         sz[x]+=sz[to[i]];
 42         if(sz[ch[x]]<sz[to[i]])ch[x]=to[i];
 43     }
 44 }
 45 void dfs2(int x,int tp){
 46     top[x]=tp;
 47     if(ch[x])dfs2(ch[x],tp);
 48     for(int i=h[x];i;i=nx[i]){
 49         if(to[i]==ch[x]||to[i]==fa[x])continue;
 50         dfs2(to[i],to[i]);
 51     }
 52 }
 53 void dfs3(int x,int pr){
 54     dfn[x]=++dfs_clock;
 55     int now=stack_top;
 56     for(int i=h[x];i;i=nx[i]){
 57         if(to[i]==pr)continue;
 58         dfs3(to[i],x);
 59         if(stack_top-now>=block_size){
 60             block_count++;
 61             while(now!=stack_top){
 62                 pos[stack[stack_top--]]=block_count;
 63             }   
 64         }
 65     }
 66     stack[++stack_top]=x;
 67 }
 68 int lca(int u,int v){
 69     while(top[u]!=top[v])dep[top[u]]>dep[top[v]]?u=fa[top[u]]:v=fa[top[v]];
 70     return dep[u]<dep[v]?u:v;
 71 }
 72 void reverse(int x){
 73     if(vis[x])ans-=W[num[C[x]]]*V[C[x]],num[C[x]]--;
 74     else num[C[x]]++,ans+=W[num[C[x]]]*V[C[x]];
 75     vis[x]^=1;
 76 }
 77 void change(int x,int y){
 78     if(vis[x]){reverse(x);C[x]=y;reverse(x);}
 79     else C[x]=y;
 80 }
 81 void solve(int x,int y){
 82     while(x!=y) {
 83         if(dep[x]>dep[y])reverse(x),x=fa[x];
 84         else reverse(y),y=fa[y];
 85     }
 86 }
 87 int main(){
 88     n=read();m=read();q=read();
 89     block_size = pow(n,2.0/3)*0.5;
 90     for(int i=1;i<=m;i++)V[i]=read();
 91     for(int i=1;i<=n;i++)W[i]=read();
 92     for(int i=1;i<n;i++)add_edge(read(),read());
 93     for(int i=1;i<=n;i++)pre[i]=C[i]=read();
 94     dfs1(1,0);dfs2(1,1);dfs3(1,0);
 95     block_count++;
 96     while(stack_top)pos[stack[stack_top--]]=block_count;
 97 //  for(int i=1;i<=n;i++)printf("%d ",pos[i]);
 98 //  puts("\nfsdfsdds\n");
 99     int c1=0,c2=0;
100     for(int i=1;i<=q;i++)
101     {
102         int typ=read(),x=read(),y=read();
103         if(!typ)
104         {
105             c1++;
106             c[c1].x=x;c[c1].y=y;c[c1].pre=pre[x];pre[x]=y;
107         }
108         else
109         {
110             c2++;
111             if(dfn[x]>dfn[y])std::swap(x,y);
112             b[c2].x=x;b[c2].y=y;b[c2].id=c2;b[c2].t=c1;
113         }
114     }
115     std::sort(b+1,b+c2+1);
116     for(int i=1;i<=b[1].t;i++)
117         change(c[i].x,c[i].y);
118     solve(b[1].x,b[1].y);
119     int t=lca(b[1].x,b[1].y);
120     reverse(t);res[b[1].id]=ans;reverse(t);
121     for(int i=2;i<=c2;i++)
122     {
123         for(int j=b[i-1].t+1;j<=b[i].t;j++)
124             change(c[j].x,c[j].y);
125         for(int j=b[i-1].t;j>b[i].t;j--)
126             change(c[j].x,c[j].pre);
127         solve(b[i-1].x,b[i].x);
128         solve(b[i-1].y,b[i].y);
129         int t=lca(b[i].x,b[i].y);
130         reverse(t);res[b[i].id]=ans;reverse(t);
131     }
132     for(int i=1;i<=c2;i++)
133         printf("%lld\n",res[i]);
134     return 0;
135 }
136