P3384 【模板】树链剖分

https://www.luogu.org/problemnew/show/P3384

树链剖分模板，需要的自取

  1 #define IO std::ios::sync_with_stdio(0);
  2 #include 
  3 #define iter ::iterator
  4 using namespace  std;
  5 typedef long long ll;
  6 typedef pairP;
  7 #define pb push_back
  8 #define se second
  9 #define fi first
 10 #define rs o<<1|1
 11 #define ls o<<1
 12 #define inf 0x3f3f3f3f
 13 const int N=1e5+5;
 14 int n,q,rt,mod;
 15 int a[N],b[N];
 16 vector<int>g[N];
 17 int deep[N],fa[N],id[N],son[N],siz[N],top[N];
 18 int sumv[N*4],add[N*4];
 19 int L[N*4],R[N*4];
 20 void dfs1(int u,int f){
 21     deep[u]=deep[f]+1;
 22     fa[u]=f;
 23     siz[u]=1;
 24     int mx=-1;
 25     for(int i=0;i){
 26         int v=g[u][i];
 27         if(v==f)continue;
 28         dfs1(v,u);
 29         siz[u]+=siz[v];
 30         if(siz[v]>mx){
 31             mx=siz[v];
 32             son[u]=v;
 33         }
 34     }
 35 }
 36 int tim;
 37 void dfs2(int u,int topf){
 38     id[u]=++tim;
 39     a[tim]=b[u];
 40     top[u]=topf;
 41     if(!son[u])return;
 42     dfs2(son[u],topf);
 43     for(int i=0;i){
 44         int v=g[u][i];
 45         if(v==fa[u]||v==son[u])continue;
 46         dfs2(v,v);
 47     }
 48 }
 49 void push(int o){
 50     sumv[o]=(sumv[ls]+sumv[rs])%mod;
 51 }
 52 void down(int o){
 53     if(add[o]){
 54         add[o]%=mod;
 55         sumv[ls]+=add[o]*(R[ls]-L[ls]+1);
 56         sumv[ls]%=mod;
 57         sumv[rs]+=add[o]*(R[rs]-L[rs]+1);
 58         sumv[rs]%=mod;
 59         add[ls]+=add[o];
 60         add[ls]%=mod;
 61         add[rs]+=add[o];
 62         add[rs]%=mod;
 63         add[o]=0;
 64     }
 65 }
 66 void build(int o,int l,int r){
 67     if(l==r){
 68         sumv[o]=a[l]%mod;
 69         L[o]=R[o]=l;
 70         return;
 71     }
 72     int m=(l+r)/2;
 73     build(ls,l,m);
 74     build(rs,m+1,r);
 75     L[o]=L[ls];
 76     R[o]=R[rs];
 77     push(o);
 78 }
 79 int qu(int o,int l,int r,int ql,int qr){
 80     if(l>=ql&&r<=qr){
 81         return sumv[o]%mod;
 82     }
 83     down(o);
 84     int m=(l+r)/2;
 85     int res=0;
 86     if(ql<=m)res+=qu(ls,l,m,ql,qr)%mod;
 87     res%=mod;
 88     if(qr>m)res+=qu(rs,m+1,r,ql,qr)%mod;
 89     res%=mod;
 90     return res;
 91 }
 92 void up(int o,int l,int r,int ql,int qr,int v){
 93     if(l>=ql&&r<=qr){
 94         v%=mod;
 95         add[o]+=v;
 96         add[o]%=mod;
 97         sumv[o]+=v*(r-l+1);
 98         sumv[o]%=mod;
 99         return;
100     }
101     down(o);
102     int m=(l+r)/2;
103     if(ql<=m)up(ls,l,m,ql,qr,v);
104     if(qr>m)up(rs,m+1,r,ql,qr,v);
105     push(o);
106 }
107 void qu1(int x,int y){//x节点到y节点的路径和
108     int res=0;
109     while(top[x]!=top[y]){
110         if(deep[top[x]]<deep[top[y]])swap(x,y);
111         res+=qu(1,1,n,id[top[x]],id[x]);
112         res%=mod;
113         x=fa[top[x]];
114     }
115     if(deep[x]>deep[y])swap(x,y);
116     res+=qu(1,1,n,id[x],id[y])%mod;
117     res%=mod;
118     printf("%d\n",res);
119 }
120 void up1(int x,int y,int z){//把x节点到y节点的路径所有点加z
121     while(top[x]!=top[y]){
122         if(deep[top[x]]<deep[top[y]])swap(x,y);
123         up(1,1,n,id[top[x]],id[x],z);
124         x=fa[top[x]];
125     }
126     if(deep[x]>deep[y])swap(x,y);
127     up(1,1,n,id[x],id[y],z);
128 }
129 int main(){
130     scanf("%d%d%d%d",&n,&q,&rt,&mod);
131     for(int i=1;i<=n;i++){
132         scanf("%d",&b[i]);
133         b[i]%=mod;
134     }
135     for(int i=1;i){
136         int x,y;
137         scanf("%d%d",&x,&y);
138         g[x].pb(y);
139         g[y].pb(x);
140     }
141     dfs1(rt,0);
142     dfs2(rt,rt);
143     build(1,1,n);
144     while(q--){
145         int op,x,y,z;
146         scanf("%d",&op);
147         if(op==1){//把x节点到y节点的路径所有点加z
148             scanf("%d%d%d",&x,&y,&z);
149             up1(x,y,z);
150         }
151         else if(op==2){//查询x节点到y节点的路径和
152             scanf("%d%d",&x,&y);
153             qu1(x,y);
154         }
155         else if(op==3){////把x节点的所有子孙加y
156             scanf("%d%d",&x,&y);
157             up(1,1,n,id[x],id[x]+siz[x]-1,y);
158         }
159         else{//查询x节点的子孙和
160             scanf("%d",&x);
161             printf("%d\n",qu(1,1,n,id[x],id[x]+siz[x]-1)%mod);
162         }
163     }
164 }