hdu 5052 Yaoge’s maximum profit

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5052

题目描述：输入一棵树，带点权，然后有一些询问， a b c，表示从a走到b，你可以在某个节点处买东西，花费为该点的点权，你在后面的点的位置可以卖出去，收益是该点的点权，询问你能获得的最大收益。然后将a 到 b，路径上的所有点加上权值c

解题思路：kuangbin大神题解是LCT，弱渣只会树链剖分，想想还是能做的。。。只是代码有点长。o  s-h-i-t...

由于树的形态是不会改变的所以一开始就没有往LCT方面想。。。。（其实想到也做不出来。。。都是泪。。。）

然后线段树维护，区间的最大值max_val，最小值min_val，从左往右的最大收益down，从右往左的最大收益up，然后就是套一个树链剖分的模板，搞一搞，就可以了

代码实在有点长，建议先想好了再敲键盘。。。不然，只能悲剧。。。

//#pragma comment(linker,"/STACK:102400000,102400000")#include<stdio.h>#include<iostream>#include<string.h>#include<math.h>#include<algorithm>#include<vector>#include<map>#include<set>#include<queue>#include<string>#define ll long long#define db double#define PB push_back#define lson k<<1#define rson k<<1|1using namespace std;const int N = 100005;const int INF = 1000000000;const int M = N << 1;int num, head[N], to[M], next[M], nedge;void init(){    memset(head, -1, sizeof(head));    nedge = 0, num = 0;}void add(int a, int b){    to[nedge] = b, next[nedge] = head[a], head[a] = nedge++;}int son[N], tol[N], dep[N], fa[N];void get_son_dep(int k, int dp){    tol[k] = 1, son[k] = 0, dep[k] = dp;    for(int i = head[k]; i >= 0; i = next[i])    {        if(to[i] != fa[k])        {            fa[to[i]] = k;            get_son_dep(to[i], dp + 1);            tol[k] += tol[to[i]];            if(tol[to[i]] > tol[son[k]]) son[k] = to[i];        }    }}int w[N], top[N];void build(int k, int rt){    w[k] = ++num, top[k] = rt;    if(son[k] != 0) build(son[k], rt);    for(int i = head[k]; i >= 0; i = next[i])    {        if(to[i] != fa[k] && to[i] != son[k])        {            build(to[i], to[i]);        }    }}struct node{    int l, r;    int max_val, min_val, up, down, lz;} tr[N << 2];void push_up(int k){    tr[k].max_val = max(tr[lson].max_val, tr[rson].max_val);    tr[k].min_val = min(tr[lson].min_val, tr[rson].min_val);    tr[k].down = max(tr[rson].max_val - tr[lson].min_val, max(tr[lson].down, tr[rson].down));    tr[k].up=max(tr[lson].max_val-tr[rson].min_val,max(tr[lson].up,tr[rson].up));}int pt[N];int v[N];void build_tr(int k, int l, int r){    tr[k].l = l, tr[k].r = r;    tr[k].lz = 0;    if(l < r)    {        int mid = (l + r) >> 1;        build_tr(lson, l, mid);        build_tr(rson, mid + 1, r);        push_up(k);    }    else tr[k].max_val = tr[k].min_val = v[pt[l]], tr[k].up=tr[k].down = 0;}void push_down(int k){    if(tr[k].lz)    {        tr[lson].lz += tr[k].lz, tr[rson].lz += tr[k].lz;        tr[lson].max_val += tr[k].lz, tr[rson].max_val += tr[k].lz;        tr[lson].min_val += tr[k].lz, tr[rson].min_val += tr[k].lz;        tr[k].lz = 0;    }}void add(int k,int l,int r,int val){    if(tr[k].l==l&&tr[k].r==r)    {        tr[k].lz+=val;        tr[k].max_val+=val;        tr[k].min_val+=val;        return;    }    push_down(k);    int mid=(tr[k].l+tr[k].r)>>1;    if(r<=mid) add(lson,l,r,val);    else if(l>mid) add(rson,l,r,val);    else add(lson,l,mid,val),add(rson,mid+1,r,val);    push_up(k);}int maxval,minval,pfct;   //-INF INF 0pair<int,int> query(int k,int l,int r,bool fl){    if(tr[k].l==l&&tr[k].r==r)    {        maxval=max(maxval,tr[k].max_val);        minval=min(minval,tr[k].min_val);        if(fl) pfct=max(pfct,tr[k].up);        else pfct=max(pfct,tr[k].down);        return make_pair(tr[k].max_val,tr[k].min_val);    }    push_down(k);    int mid=(tr[k].l+tr[k].r)>>1;    if(r<=mid) return query(lson,l,r,fl);    else if(l>mid) return query(rson,l,r,fl);    else    {        pair<int,int> t1=query(lson,l,mid,fl);        pair<int,int> t2=query(rson,mid+1,r,fl);        if(fl) pfct=max(pfct,t1.first-t2.second);        else pfct=max(pfct,t2.first-t1.second);        return make_pair(max(t1.first,t2.first),min(t1.second,t2.second));    }}struct P{    int max_val, min_val, pfct, id;    bool operator < (const P &t) const    {        return id < t.id;    }} seq[1000];int mval[1000];int lseq;int find_ans(int a, int b, int val){    lseq = 0;    int tpa = top[a], tpb = top[b];    int la = 0, lb = INF;    while(tpa != tpb)    {        maxval = -INF, minval = INF, pfct = 0;        if(dep[tpa] < dep[tpb])        {            query(1, w[tpb], w[b],false);            add(1, w[tpb], w[b], val);            b = fa[tpb], tpb = top[b];            seq[lseq].id = lb--;        }        else        {            query(1, w[tpa], w[a],true);            add(1, w[tpa], w[a], val);            a = fa[tpa], tpa = top[a];            seq[lseq].id = la++;        }        seq[lseq].max_val = maxval, seq[lseq].min_val = minval, seq[lseq].pfct = pfct;        lseq++;    }//    if(a!=b){        maxval = -INF, minval = INF, pfct = 0;        if(dep[a] < dep[b])        {            query(1, w[a], w[b],false);  //            add(1, w[a], w[b], val);  //            b = fa[tpb], tpb = top[b];            seq[lseq].id = lb--;        }        else        {            query(1, w[b], w[a],true);  //            add(1, w[b], w[a], val);  //            a = fa[tpa], tpa = top[a];            seq[lseq].id = la++;        }        seq[lseq].max_val = maxval, seq[lseq].min_val = minval, seq[lseq].pfct = pfct;        lseq++;//    }    sort(seq,seq+lseq);//    for(int i=0;i<lseq;i++) printf("%d %d\n",seq[i].max_val,seq[i].min_val);    int res=0;    mval[lseq]=0;    for(int i=lseq-1;i>=0;i--)    {        mval[i]=max(mval[i+1],seq[i].max_val);        res=max(res,seq[i].pfct);    }    int mival=INF;    for(int i=0;i<lseq-1;i++)    {        mival=min(mival,seq[i].min_val);        res=max(res,mval[i+1]-mival);    }    return res;}void getint(int &x){    char c = getchar();    bool sg = false;    while(!(c >= '0' && c <= '9' || c == '-')) c = getchar();    if(c == '-') sg = true;    x = c - '0';    while((c = getchar()) >= '0' && c <= '9') x = x * 10 + c - '0';    sg&&(x = -x);}int main(){#ifdef PKWV    freopen("in.in", "r", stdin);#endif // PKWV    int T, cas = 1;    getint(T);    while(T--)    {        int n;        getint(n);        for(int i=1;i<=n;i++) getint(v[i]);        init();        for(int i=1;i<n;i++)        {            int a,b;            getint(a),getint(b);            add(a,b),add(b,a);        }        int rt=n/2+1;        fa[rt]=0;        get_son_dep(rt,0);        build(rt,rt);        for(int i=1;i<=n;i++) pt[w[i]]=i;        build_tr(1,1,num);        int Q;        scanf("%d",&Q);        while(Q--)        {            int a,b,c;            getint(a),getint(b),getint(c);            printf("%d\n",find_ans(a,b,c));        }    }    return 0;}