Fzu 2082 tree Chain split

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Test instructions: Chinese

Idea: The most basic tree chain, interval summation and single-point update, the result of a long long is nothing

#include <vector> #include <stdio.h> #include <string.h> #include <stdlib.h> #include < Iostream> #include <algorithm>using namespace std;typedef long long ll;typedef unsigned long long ull;const int in f=0x3f3f3f3f;const ll Inf=0x3f3f3f3f3f3f3f3fll;const int Maxn=100010;int fa[maxn],siz[maxn],son[maxn],w[maxn],p[ MAXN],DEP[MAXN],FP[MAXN],HEAD[MAXN];//FA is the parent node, Siz is the largest siz in the child node, DEP is the depth, son is heavy, and W represents the position in the segment tree in int num[maxn<<2]; int tree_id,n,kkk=0;struct node{int to,next;} Ee[maxn*10];void Add_edge (int u,int v) {ee[kkk].to=v; ee[kkk].next=head[u];head[u]=kkk++;}    void dfs1 (int u,int ff,int deep) {son[u]=0;fa[u]=ff;siz[u]=1;dep[u]=deep;        for (int i=head[u];i!=-1;i=ee[i].next) {int v=ee[i].to;        if (V==FF) continue;        DFS1 (v,u,deep+1);        SIZ[U]+=SIZ[V];    if (Siz[v]>siz[son[u]]) son[u]=v;    }}void DFS2 (int u,int ff) {w[u]=++tree_id;p[u]=ff;    if (Son[u]) DFS2 (SON[U],FF);    else return; for (int i=head[u];i!=-1;i=ee[i].next) {        int v=ee[i].to;    if (V!=fa[u]&&v!=son[u]) DFS2 (V,V);        }}void Update (int pos,int val,int le,int ri,int node) {if (Le==ri) {num[node]=val;    return;    } int t= (LE+RI) >>1;    if (pos<=t) update (POS,VAL,LE,T,NODE&LT;&LT;1);    else update (POS,VAL,T+1,RI,NODE&LT;&LT;1|1); NUM[NODE]=NUM[NODE&LT;&LT;1]+NUM[NODE&LT;&LT;1|1];}    int query (int l,int r,int le,int ri,int node) {if (l<=le&&ri<=r) return Num[node];    int t= (LE+RI) >>1,ans=0;    if (l<=t) ans=ans+query (l,r,le,t,node<<1);    if (r>t) ans=ans+query (l,r,t+1,ri,node<<1|1); return ans;}    ll getmin (int u,int v) {int f1=p[u],f2=p[v];    ll Tmp=0;            while (F1!=F2) {if (Dep[f1]<dep[f2]) {swap (F1,F2);        Swap (U,V);        } tmp=tmp+query (w[f1],w[u],1,n,1);    U=fa[f1];f1=p[u];    } if (U==V) return tmp;    if (Dep[u]>dep[v]) swap (U,V); Return Tmp+query (w[son[u]],w[v],1,n,1);} int U[maxn],v[maxn],c[maxn];int mAin () {int u,v,q,op;        while (scanf ("%d%d", &n,&q)!=-1) {memset (head,-1,sizeof (head));        memset (son,0,sizeof (son)); tree_id=0;kkk=0;            for (int i=0;i<n-1;i++) {scanf ("%d%d%d", &u[i],&v[i],&c[i]);        Add_edge (U[i],v[i]); Add_edge (V[i],u[i]);        } DFS1 (1,1,0);        DFS2 (a);        memset (num,0,sizeof (num));            for (int i=0;i<n-1;i++) {if (Dep[u[i]]>dep[v[i]) swap (u[i],v[i]);        Update (w[v[i]],c[i],1,n,1);            } while (q--) {scanf ("%d%d%d", &op,&u,&v);            if (op==0) update (w[v[u-1]],v,1,n,1);                else{ll Fans=getmin (u,v);            printf ("%i64d\n", fans); }}} return 0;}

Fzu 2082 tree Chain split