HDU 3966 树链剖分

来源：互联网发布：sql培训机构编辑：程序博客网时间：2024/05/16 05:09

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题意：给一个树，三个操作，一个询问点的权值，一个将u到v路径上的点的权值增加，一个是减少

思路：还是比较裸的模版题，线段树改为区间更新就好了

#pragma comment(linker, "/STACK:1024000000,1024000000")#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 inf=0x3f3f3f3f;const ll INF=0x3f3f3f3f3f3f3f3fll;const int maxn=50010;int fa[maxn],siz[maxn],son[maxn],w[maxn],p[maxn],dep[maxn],fp[maxn];//fa为父节点，siz为子节点中siz最大的，dep为深度，son为重儿子，w表示在线段树中的位置int num[maxn<<2],lazy[maxn<<2];int tree_id,n;vector<int>G[maxn];void dfs1(int u,int ff,int deep){    son[u]=0;fa[u]=ff;siz[u]=1;dep[u]=deep;    for(unsigned int i=0;i<G[u].size();i++){        int v=G[u][i];        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(unsigned int i=0;i<G[u].size();i++){        int v=G[u][i];        if(v!=fa[u]&&v!=son[u]) dfs2(v,v);    }}void pushdown(int node){    if(lazy[node]){        lazy[node<<1]+=lazy[node];        lazy[node<<1|1]+=lazy[node];        num[node<<1]+=lazy[node];        num[node<<1|1]+=lazy[node];        lazy[node]=0;    }}void update(int l,int r,int val,int le,int ri,int node){    if(l<=le&&ri<=r){        num[node]+=val;        lazy[node]+=val;        return ;    }    pushdown(node);    int t=(le+ri)>>1;    if(l<=t) update(l,r,val,le,t,node<<1);    if(r>t) update(l,r,val,t+1,ri,node<<1|1);}int query(int pos,int le,int ri,int node){    if(le==ri) return num[node];    pushdown(node);    int t=(le+ri)>>1;    if(pos<=t) query(pos,le,t,node<<1);    else query(pos,t+1,ri,node<<1|1);}void getans(int u,int v,int val){    int f1=p[u],f2=p[v];    while(f1!=f2){        if(dep[f1]<dep[f2]){            swap(f1,f2);            swap(u,v);        }        update(w[f1],w[u],val,1,n,1);        u=fa[f1];f1=p[u];    }    if(dep[u]>dep[v]) swap(u,v);    update(w[u],w[v],val,1,n,1);}int A[maxn];int main(){    char str[10];    int m,q,u,v;    while(scanf("%d%d%d",&n,&m,&q)!=-1){        for(int i=1;i<=n;i++) scanf("%d",&A[i]);        for(int i=0;i<maxn;i++) G[i].clear();        memset(son,0,sizeof(son));tree_id=0;        memset(lazy,0,sizeof(lazy));        for(int i=0;i<m;i++){            scanf("%d%d",&u,&v);            G[u].push_back(v);            G[v].push_back(u);        }        dfs1(1,1,0);        dfs2(1,1);        memset(num,0,sizeof(num));        while(q--){            scanf("%s",str);            if(str[0]=='Q'){                scanf("%d",&u);                printf("%d\n",A[u]+query(w[u],1,n,1));            }else if(str[0]=='I'){                scanf("%d%d%d",&u,&v,&m);                getans(u,v,m);            }else{                scanf("%d%d%d",&u,&v,&m);                getans(u,v,-m);            }        }    }    return 0;}

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