省选D2T3考板子可真是不多见呢。。。。~~~
这题就是一个裸的树链剖分,对于每一个Add操作,维护从u至v的路径,对于每一个Query操作,询问以u为根的子树之和。如果不会树链剖分可以看我的往期博客,具体细节在代码之中就不多赘述了~
下面给出参考代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<iostream> 2 #include<cstdio> 3 #define N 400005 4 #define M 800005 5 #define lc k*2 6 #define rc k*2+1 7 #define mid (l+r)/2 8 #define int long long 9 using namespace std; 10 struct node 11 { 12 int l,r,w,tag; 13 }tree[4*N]; 14 int n,m,r,p,x,y,z; 15 int v[M],head[M],nxt[M],cnt; 16 int dep[N],fa[N],son[N],size[N],top[N],seg[N],id; 17 char q; 18 int read() 19 { 20 int x=0,f=1;char ch=getchar(); 21 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 22 while(ch>='0'&&ch<='9'){x=x*10+ch-48;ch=getchar();} 23 return x*f; 24 } 25 void add(int a,int b) 26 { 27 v[++cnt]=b; 28 nxt[cnt]=head[a]; 29 head[a]=cnt; 30 } 31 void dfs1(int node,int father) 32 { 33 dep[node]=dep[father]+1; 34 fa[node]=father; 35 size[node]=1; 36 int maxson=-1; 37 for(int i=head[node];i;i=nxt[i]) 38 { 39 int go=v[i]; 40 if(go==father)continue; 41 dfs1(go,node); 42 size[node]+=size[go]; 43 if(size[go]>maxson)maxson=size[go],son[node]=go; 44 } 45 } 46 void dfs2(int node,int topfather) 47 { 48 seg[node]=++id; 49 top[node]=topfather; 50 if(!son[node])return; 51 dfs2(son[node],topfather); 52 for(int i=head[node];i;i=nxt[i]) 53 { 54 int go=v[i]; 55 if(go==fa[node]||go==son[node])continue; 56 dfs2(go,go); 57 } 58 } 59 void build(int l,int r,int k) 60 { 61 tree[k].l=l;tree[k].r=r; 62 if(l==r)return; 63 build(l,mid,lc); 64 build(mid+1,r,rc); 65 } 66 void pushdown(int k) 67 { 68 tree[lc].tag+=tree[k].tag; 69 tree[rc].tag+=tree[k].tag; 70 tree[lc].w+=(tree[lc].r-tree[lc].l+1)*tree[k].tag; 71 tree[rc].w+=(tree[rc].r-tree[rc].l+1)*tree[k].tag; 72 tree[k].tag=0; 73 return; 74 } 75 void pushup(int k) 76 { 77 tree[k].w=tree[lc].w+tree[rc].w; 78 return; 79 } 80 void add(int x,int y,int k) 81 { 82 int l=tree[k].l,r=tree[k].r; 83 if(l>=x&&r<=y) 84 { 85 tree[k].tag+=z; 86 tree[k].w+=(r-l+1)*z; 87 return; 88 } 89 if(tree[k].tag)pushdown(k); 90 if(x<=mid)add(x,y,lc); 91 if(y>mid)add(x,y,rc); 92 pushup(k); 93 } 94 int query(int x,int y,int k) 95 { 96 int l=tree[k].l,r=tree[k].r; 97 if(l>=x&&r<=y) 98 { 99 return tree[k].w; 100 } 101 if(tree[k].tag)pushdown(k); 102 int res=0; 103 if(x<=mid)res+=query(x,y,lc); 104 if(y>mid)res+=query(x,y,rc); 105 return res; 106 } 107 void Tadd(int x,int y) 108 { 109 while(top[x]!=top[y]) 110 { 111 if(dep[top[x]]<dep[top[y]])swap(x,y); 112 add(seg[top[x]],seg[x],1); 113 x=fa[top[x]]; 114 } 115 if(dep[x]>dep[y])swap(x,y); 116 add(seg[x],seg[y],1); 117 } 118 int Treequery(int x) 119 { 120 return query(seg[x],seg[x]+size[x]-1,1); 121 } 122 signed main() 123 { 124 n=read(); 125 for(int i=1;i<n;i++) 126 { 127 x=read();y=read(); 128 x++;y++; 129 add(x,y);add(y,x); 130 } 131 dfs1(1,0);dfs2(1,1); 132 build(1,n,1); 133 m=read(); 134 while(m--) 135 { 136 cin>>q; 137 if(q=='A') 138 { 139 x=read();y=read();z=read(); 140 x++;y++; 141 Tadd(x,y); 142 } 143 else 144 { 145 x=read(); 146 x++; 147 cout<<Treequery(x)<<endl; 148 } 149 } 150 }