[JSOI2008] [BZOJ1036] Tree statistics-tree chain split

Description

There are n nodes on a tree, numbered 1 to N, each with a weight of W.

We will ask you to do some work on this tree in the following form:

I. Change u t: Changing the weight of the node U to t

Ii. QMAX u V: ask for the maximum weight of the node on the path from point u to V

Iii. qsum u V: ask for the weights and values of the nodes on the path from point u to V

Note: Nodes on the path from point u to v include U and V itself

Input & Outputinput

The first behavior of the input file is an integer n, which represents the number of nodes.

The next n–1 line, with 2 integers a and b per line, indicates that there is an edge connected between Node A and Node B.

The next line is n integers, and the I integer WI indicates the weight of node I.

The next 1 lines, an integer q, represent the total number of operations.

The next Q line, one operation per line, is given in the form "Change U T" or "QMAX u V" or "Qsum u V".

Output

For each "QMAX" or "qsum" operation, each line outputs an integer representing the result of the required output.

Sampleinput

41 22 34 14 2 1 312QMAX 3 4QMAX 3 3QMAX 3 2QMAX 2 3QSUM 3 4QSUM 2 1CHANGE 1 5QMAX 3 4CHANGE 3 6QMAX 3 4QMAX 2 4QSUM 3 4

Output

412210656516

Solution

Porcelain single point modification, the path query maximum value, the path query point right and, compared to bare tree profile.
Fear is not a constant more than the Rokua (escape
Code:

#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include < cctype> #include <string>using std::min;using std::max;using std::swap;using std::isdigit;using std::memset;  Using std::string;using std::cin;using std::cout;using std::endl;using std::ios;const int maxn = 30005;struct edge{int TO,NXT;}    e[maxn<<1];struct node{int W,mx,tag;    Node () {mx = w = tag = 0; }}TR[MAXN << 2];int lnk[maxn],edgenum;int top[maxn],size[maxn],dep[maxn],son[maxn],f[maxn],dfn[maxn],dfn2[    Maxn];int cnt;int w[maxn],v[maxn],n,m,q;void Add (int bgn,int end) {e[++edgenum].to = end;    E[EDGENUM].NXT = LNK[BGN]; LNK[BGN] = edgenum;}    void Dfs (int x,int fa,int d) {size[x] = 1;    F[X] = FA;    DEP[X] = D;        for (int p = lnk[x]; p; p = e[p].nxt) {int y = e[p].to;        if (y = = FA) continue;        DFS (y, x, D + 1);        SIZE[X] + = Size[y];    if (Size[y] > Size[son[x]]) son[x] = y; }}void DFS2 (int x,int init) {dfn[x] = ++cnt;    W[CNT] = v[x];    TOP[X] = init;    if (!son[x]) return;    DFS2 (Son[x],init);        for (int p = lnk[x]; p; p = e[p].nxt) {int y = e[p].to; if (y = = f[x]| |        y = = Son[x]) continue;    DFS2 (Y,y);        }}void Build (int cur,int l,int R) {if (L = = r) {TR[CUR].W = w[l];        tr[cur].mx = W[l];    Return    } int mid = (L + r) >> 1;    Build (Cur<<1,l,mid);    Build (Cur<<1|1,mid + 1,r);    TR[CUR].W = TR[CUR&LT;&LT;1].W + tr[cur<<1|1].w; tr[cur].mx = max (tr[cur<<1].mx,tr[cur<<1|1].mx);}    int querys (int cur,int l,int r,int l,int r) {int res = 0;        if (l <= l && R >= R) {res + = TR[CUR].W;    return res;    } int mid = (l+r) >>1;    if (l<=mid) res + = Querys (cur<<1,l,mid,l,r);    if (r>mid) res + = Querys (cur<<1|1,mid+1,r,l,r); return res;}    int Queryx (int cur,int l,int r,int l,int r) {int res =-214748; if (l <= l && R; = r) return tr[cur].mx;    int mid = (l+r) >>1;    if (l<=mid) res = max (res, Queryx (cur<<1,l,mid,l,r));    if (r>mid) res = max (res, Queryx (cur<<1|1,mid+1,r,l,r)); return res;}        void pupdate (int cur,int l,int r,int p,int c) {if (L = = r) {TR[CUR].W = C;        tr[cur].mx = C;    Return        } else{int mid = (l+r) >>1;        if (P <= mid) pupdate (CUR&LT;&LT;1,L,MID,P,C);        else Pupdate (CUR&LT;&LT;1|1,MID+1,R,P,C);        TR[CUR].W = TR[CUR&LT;&LT;1].W + tr[cur<<1|1].w;    tr[cur].mx = max (tr[cur<<1].mx,tr[cur<<1|1].mx);    }}//---------------------------int queryts (int x,int y) {int ans = 0;        while (top[x]! = Top[y]) {if (Dep[top[x]] < Dep[top[y]]) swap (x, y);        Ans + = Querys (1,1,n,dfn[top[x]],dfn[x]);    x = f[top[x]];    } if (Dep[x]>dep[y]) swap (x, y);    Ans + = Querys (1,1,n,dfn[x],dfn[y]); return ans;}    int querytx (int x,int y) {int ans =-214748; while (top[x]! = Top[y]) {if (Dep[top[x]] < dep[top[y]) swap (x, y);        ans = max (ans, Queryx (1,1,n,dfn[top[x]],dfn[x));    x = f[top[x]];    } if (Dep[x] > Dep[y]) swap (x, y);    ans = max (ans, Queryx (1,1,n,dfn[x],dfn[y)); return ans;}    int main () {Ios::sync_with_stdio (false);    string S;    int x, y;    CIN >> N;        for (int i = 1; i < n; ++i) {cin >> x >> y;        Add (x, y);    Add (y,x);    } for (int i = 1; I <= n; ++i) cin >> V[i];    DFS (1,0,1);    DFS2 (a);    Build (1,1,n);    CIN >> Q;        for (int i = 1; I <= Q; ++i) {cin >> S;            if (s[0] = = ' C ') {cin >> x >> y;        Pupdate (1,1,n,dfn[x],y);            } else if (s[0] = = ' Q ' && s[1] = = ' M ') {cin >> x >> y;        cout << Querytx (x, y) << Endl;            } else {cin >> x >> y; cout << QuEryts (x, y) << Endl; }} return 0;}

[jsoi2008][bzoj1036] Tree statistics-tree chain split