HDU3966 树链剖分

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题目：HDU3966

题意：一棵树上每个节点有一些值，三个操作（其实是俩）
操作I是增加一条链所有点的值
操作D是减少一条链所有点的值
操作Q是查询某个点的值
网上说得手工扩栈，我没扩栈也A掉了

//#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cstdio>#include <iostream>#include <cstring>#include <vector>using namespace std;#define REP(I,N) for (I=0;I<N;I++)#define rREP(I,N) for (I=N-1;I>=0;I--)#define rep(I,S,N) for (I=S;I<N;I++)#define rrep(I,S,N) for (I=N-1;I>=S;I--)#define FOR(I,S,N) for (I=S;I<=N;I++)typedef unsigned long long ULL;typedef long long LL;const int INF=0x3f3f3f3f;const LL INFF=0x3f3f3f3f3f3f3f3fll;const LL M=1e9+7;const LL maxn=50007;const double eps=0.00000001;LL gcd(LL a,LL b){return b?gcd(b,a%b):a;}template<typename T>inline T abs(T a,T b) {return a>0?a:-a;}int lazy[maxn*4],a[maxn];//用lazy就好吧。。这个不会pushupvoid pushdown(int x){    if (lazy[x]){        lazy[x<<1]+=lazy[x];        lazy[x<<1|1]+=lazy[x];        lazy[x]=0;    }}void build(int x,int l,int r){    lazy[x]=0;    if (l==r) {        lazy[x]=a[l];        return;    }    int mid=(l+r)/2;    build(x<<1,l,mid);    build(x<<1|1,mid+1,r);}void update(int x,int l,int r,int val,int L,int R){    if (l<=L&&R<=r){        lazy[x]+=val;        return;    }    pushdown(x);    int mid=(L+R)/2;    if (mid>=l) update(x<<1,l,r,val,L,mid);    if (r>mid) update(x<<1|1,l,r,val,mid+1,R);}int query(int x,int pos,int L,int R){    if (L==R) return lazy[x];    pushdown(x);    int mid=(L+R)/2;    if (mid>=pos) return query(x<<1,pos,L,mid);    return query(x<<1|1,pos,mid+1,R);}int n,q,s,T;int i,j,k;int u,v,len;vector<int> edge[maxn];int fa[maxn],son[maxn],sz[maxn],top[maxn],id[maxn],dep[maxn];int b[maxn];int tot;void dfs1(int u,int from,int depth){    int v,i,mx=-1;    sz[u]=1;fa[u]=from;dep[u]=depth;son[u]=0;    REP(i,edge[u].size()){        v=edge[u][i];        if (v==from) continue;        dfs1(v,u,depth+1);        sz[u]+=sz[v];        if (sz[v]>mx) mx=sz[v],son[u]=v;    }}void dfs2(int u,int x){    int v,i;    top[u]=x;id[u]=++tot;    if (son[u]) dfs2(son[u],x);    REP(i,edge[u].size()){        v=edge[u][i];        if (v==fa[u]||v==son[u]) continue;        dfs2(v,v);    }}void Change(int x,int y,int val){    while (top[x]!=top[y]){        if (dep[top[x]]<dep[top[y]]) swap(x,y);        update(1,id[top[x]],id[x],val,1,tot);        x=fa[top[x]];    }    if (dep[x]>dep[y]) swap(x,y);    update(1,id[x],id[y],val,1,tot);}int main(){    while (~scanf("%d%d%d",&n,&i,&q)){        FOR(i,1,n) scanf("%d",&b[i]);        FOR(i,1,n) edge[i].clear();        REP(i,n-1){            scanf("%d%d",&u,&v);            edge[u].push_back(v);            edge[v].push_back(u);        }        tot=0;        dfs1(1,0,1);        dfs2(1,1);        FOR(i,1,n) a[id[i]]=b[i];//映射一下        build(1,1,tot);        REP(i,q){            char c;            int c1,c2,k;            c=' ';            while (c!='I'&&c!='D'&&c!='Q') c=getchar();            if (c=='I'){                scanf("%d%d%d",&c1,&c2,&k);                Change(c1,c2,k);            }            if (c=='D'){                scanf("%d%d%d",&c1,&c2,&k);                Change(c1,c2,-k);            }            if (c=='Q'){                scanf("%d",&c1);                printf("%d\n",query(1,id[c1],1,tot));            }        }    }}/**/

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