Hdu 5052 Yaoge’s maximum profit 动态树 LCT link-cut tree

Yaoge’s maximum profit

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 520    Accepted Submission(s): 152

Problem Description

Yaoge likes to eat chicken chops late at night. Yaoge has eaten too many chicken chops, so that Yaoge knows the pattern in the world of chicken chops. There are N cities in the world numbered from 1 to N . There are some roads between some cities, and there is one and only one simple path between each pair of cities, i.e. the cities are connected like a tree. When Yaoge moves along a path, Yaoge can choose one city to buy ONE chicken chop and sell it in a city after the city Yaoge buy it. So Yaoge can get profit if Yaoge sell the chicken chop with higher price. Yaoge is famous in the world. AFTER Yaoge has completed one travel, the price of the chicken chop in each city on that travel path will be increased by V .

 

Input

The first line contains an integer T (0 < T ≤ 10), the number of test cases you need to solve. For each test case, the first line contains an integer N (0 < N ≤ 50000), the number of cities. For each of the next N lines, the i-th line contains an integer W_i(0 < W_i ≤ 10000), the price of the chicken chop in city i. Each of the next N - 1 lines contains two integers X Y (1 ≤ X, Y ≤ N ), describing a road between city X and city Y . The next line contains an integer Q(0 ≤ Q ≤ 50000), the number of queries. Each of the next Q lines contains three integer X Y V(1 ≤ X, Y ≤ N ; 0 < V ≤ 10000), meaning that Yaoge moves along the path from city X to city Y , and the price of the chicken chop in each city on the path will be increased by V AFTER Yaoge has completed this travel.

 

Output

For each query, output the maximum profit Yaoge can get. If no positive profit can be earned, output 0 instead.

 

Sample Input

15123451 22 33 44 551 5 15 1 11 1 25 1 11 2 1

 

Sample Output

40010

 

Source

2014 ACM/ICPC Asia Regional Shanghai Online

splay维护路径上的最大值max，最小值min，

以及从左到右能得到的最大利益l，从右到左能得到的最大利益r

举例：

根的最大值=max（左孩子最大值，自己的值，右孩子最大值）

根从左到右的最大值=max（左孩子的左到右最大值，右孩子的左到右最大值，max（右孩子最大值，根）-min（根，左孩子最小值））

在be_root的时候，要记得把r，l反转即可

time：2031ms

#pragma comment(linker, "/STACK:102400000,102400000")#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>using namespace std;#define maxn 500007#define inf  1000000000struct Node{    Node *fa,*ch[2];    bool rev,root;    int val,add,max,min,l,r;};Node pool[maxn];Node *nil,*tree[maxn];int cnt = 0;void init(){    cnt = 1;    nil = tree[0] = pool;    nil -> l = 0;    nil -> r = 0;    nil -> max = -inf;    nil -> min = inf;}inline Node *newnode(int val,Node *f){    pool[cnt].fa = f;    pool[cnt].ch[0] = pool[cnt].ch[1] = nil;    pool[cnt].rev = false;    pool[cnt].root = true;    pool[cnt].val = val;    pool[cnt].add = 0;    pool[cnt].max = val;    pool[cnt].min = val;    pool[cnt].l = 0;    pool[cnt].r = 0;    return &pool[cnt++];}//splay向上更新信息void update(Node *x){    x->max = x->min = x->val;    if(x->ch[0] != nil){        x->max < x->ch[0]->max?x->max=x->ch[0]->max:0;        x->min > x->ch[0]->min?x->min=x->ch[0]->min:0;    }    if(x->ch[1] != nil){        x->max < x->ch[1]->max?x->max=x->ch[1]->max:0;        x->min > x->ch[1]->min?x->min=x->ch[1]->min:0;    }    x->l = x->ch[1]->l < x->ch[0]->l?x->ch[0]->l:x->ch[1]->l;    x->l = max(x->l,max(x->ch[1]->max,x->val)                -min(x->ch[0]->min,x->val));    x->r = x->ch[1]->r < x->ch[0]->r?x->ch[0]->r:x->ch[1]->r;    x->r = max(x->r,max(x->ch[0]->max,x->val)                -min(x->ch[1]->min,x->val));}void update_rev(Node *x){    if(x == nil) return ;    x->rev = !x->rev;    swap(x->ch[0],x->ch[1]);    swap(x->l,x->r);}void update_add(Node *x,int n){    if(x == nil) return ;    x->max += n ;    x->min += n;    x->val += n;    x->add += n;}//splay下推信息void pushdown(Node *x){    if(x->add != 0){        update_add(x->ch[0],x->add);        update_add(x->ch[1],x->add);        x->add = 0;    }    if(x->rev != false){        update_rev(x->ch[0]);        update_rev(x->ch[1]);        x->rev = false;    }}//splay在root-->x的路径下推信息void push(Node *x){    if(!x->root) push(x->fa);    pushdown(x);}//将结点x旋转至splay中父亲的位置void rotate(Node *x){    Node *f = x->fa, *ff = f->fa;    int t = (f->ch[1] == x);    if(f->root) x->root = true, f->root = false;    else ff->ch[ff->ch[1] == f] = x;    x->fa = ff;    f->ch[t] = x->ch[t^1];    x->ch[t^1]->fa = f;    x->ch[t^1] = f;    f->fa = x;    update(f);}//将结点x旋转至x所在splay的根位置void splay(Node *x){    push(x);    Node *f, *ff;    while(!x->root){        f = x->fa,ff = f->fa;        if(!f->root)            if((ff->ch[1] == f) && (f->ch[1] == x)) rotate(f);            else rotate(x);        rotate(x);    }    update(x);}//将x到树根的路径并成一条pathNode *access(Node *x){    Node *y = nil;    while(x != nil){        splay(x);        x->ch[1]->root = true;        (x->ch[1] = y)->root = false;        update(x);        y = x;        x = x->fa;    }    return y;}//将结点x变成树根void be_root(Node *x){    access(x);    splay(x);    update_rev(x);}struct Edge{    int v,next;};Edge edge[2*maxn];int head[maxn],ecnt;int value[maxn];void add_edge(int u,int v){    edge[ecnt].v = v;    edge[ecnt].next = head[u];    head[u] = ecnt++;    edge[ecnt].v = u;    edge[ecnt].next = head[v];    head[v] = ecnt++;}void dfs(int u,int f){    tree[u] = newnode(value[u],tree[f]);    for(int i = head[u]; i != -1;i = edge[i].next){        if(edge[i].v == f) continue;        dfs(edge[i].v,u);    }}int main(){    int t,n,w,x,y,v,q;    Node*p;    scanf("%d",&t);    while(t--){        scanf("%d",&n);        memset(head,-1,sizeof(head));        ecnt = 0;        init();        for(int i = 1;i <= n; i++)            scanf("%d",&value[i]);        for(int i = 1; i < n; i++){            scanf("%d%d",&x,&y);            add_edge(x,y);        }        dfs(1,0);        scanf("%d",&q);        while(q--){            scanf("%d%d%d",&x,&y,&w);            be_root(tree[x]);            p = access(tree[y]);            printf("%d\n",p->l);            update_add(p,w);        }    }    return 0;}