树链剖分&树状数组区间加减区间求和模板

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#pragma comment(linker,"/STACK:100000000,100000000")#include<cstdio>#include<algorithm>#include<cstring>#include<vector>using namespace std;const int N=50010;#define ll long longint n,m,p,a[N];int ti[N],idx,par[N],son[N],sz[N],dep[N],top[N];vector<int> g[N];int dt[N];ll pre[N];void Dfs1(int u,int fa,int d){    par[u]=fa,son[u]=0,sz[u]=1,dep[u]=d;    for(int i=0;i<g[u].size();i++)    {        int v=g[u][i];        if(v==fa) continue;        Dfs1(v,u,d+1);        sz[u]+=sz[v];        if(sz[v]>sz[son[u]]) son[u]=v;    }}void Add(int l,int r,int v){    for(int i=l;i<=n;i+=i&(-i))dt[i]+=v,pre[i]+=v*l;    for(int i=r+1;i<=n;i+=i&(-i))dt[i]-=v,pre[i]-=v*(r+1);}void Dfs2(int u,int tp){    ti[u]=++idx,Add(idx,idx,a[u]),top[u]=tp;    if(son[u]) Dfs2(son[u],tp);    for(int i=0;i<g[u].size();i++)    {        int v=g[u][i];        if(v==son[u]||v==par[u]) continue;        Dfs2(v,v);    }}ll Sum(int pos){    ll ans=0;    for(int i=pos;i>0;i-=i&(-i)) ans+=1LL*(pos+1)*dt[i]-pre[i];    for(int i=pos-1;i>0;i-=i&(-i)) ans-=1LL*pos*dt[i]-pre[i];    return ans;}void Update(int u,int v,int w){    while(top[u]!=top[v])    {        if(dep[top[u]]<dep[top[v]]) swap(u,v);        Add(ti[top[u]],ti[u],w);        u=par[top[u]];    }    if(dep[u]>dep[v]) swap(u,v);    Add(ti[u],ti[v],w);//因为是点权,所以ti[u]不需要+1}int main(){    while(~scanf("%d%d%d",&n,&m,&p))    {        memset(pre,0,sizeof(pre));        memset(dt,0,sizeof(dt));        idx=0;        for(int i=1;i<=n;i++)scanf("%d",&a[i]),g[i].clear();        for(int i=1;i<n;i++)        {            int u,v;            scanf("%d%d",&u,&v);            g[u].push_back(v);            g[v].push_back(u);        }        Dfs1(1,1,1);        Dfs2(1,1);        while(p--)        {            char str[4];            int u,v,w;            scanf("%s",str);            if(str[0]=='I'||str[0]=='D')            {                scanf("%d%d%d",&u,&v,&w);                if(str[0]=='D') w=-w;                Update(u,v,w);            }            else            {                scanf("%d",&u);                printf("%I64d\n",Sum(ti[u]));            }        }    }}
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