hdu 5052 树链剖分
Yaoge’s maximum profit
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 982 Accepted Submission(s): 274
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 Wi(0 < Wi ≤ 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
1
5
1
2
3
4
5
1 2
2 3
3 4
4 5
5
1 5 1
5 1 1
1 1 2
5 1 1
1 2 1
Sample Output
4
0
0
1
0
/*
hdu 5052 树链剖分(nice)
problem:
给你一个树,每次找出u->v上面的最大差值(较小值必需在较大值前面).找出后在给路径所有点加上w
solve:
首先是线段树维护差值的问题,在这里错了很久- -. 按照以前的写习惯了,并没想区间合并时候的问题...
树链剖分查找的时候,每次只能查找一条链,所以在这里也要合并(右边链Max - 左边链Min).
而且u->v的话,因为u到(u,v)的lca的节点号是逆序的(根节点较小),所以线段树要维护 左到右and右到左的差值
hhh-2016-08-22 10:53:40
*/
#pragma comment(linker,"/STACK:124000000,124000000")
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <vector>
#include <map>
#define lson i<<1
#define rson i<<1|1
#define ll long long
#define clr(a,b) memset(a,b,sizeof(a))
#define key_val ch[ch[root][1]][0]
#define inf 0x3FFFFFFFFFFFFFFFLL
using namespace std;
const int maxn = 200100;
int head[maxn],tot,pos,son[maxn];
int top[maxn],fp[maxn],fa[maxn],dep[maxn],num[maxn],p[maxn];
int n;
ll a[maxn];
ll MAX(ll a,ll b)
{
return a>b?a:b;
}
ll MIN(ll a,ll b)
{
return a>b?b:a;
}
struct Edge
{
int to,next;
} edge[maxn<<2];
void ini()
{
tot = 0,pos = 1;
clr(head,-1),clr(son,-1);
}
void add_edge(int u,int v)
{
edge[tot].to = v,edge[tot].next = head[u],head[u] = tot++;
}
void dfs1(int u,int pre,int d)
{
// cout << u << " " <<pre <<" " <<d <<endl;
dep[u] = d;
fa[u] = pre,num[u] = 1;
for(int i = head[u]; ~i; i = edge[i].next)
{
int v = edge[i].to;
if(v != pre)
{
dfs1(v,u,d+1);
num[u] += num[v];
if(son[u] == -1 || num[v] > num[son[u]])
son[u] = v;
}
}
}
void getpos(int u,int sp)
{
top[u] = sp;
p[u] = pos++;
fp[p[u]] = u;
if(son[u] == -1)return ;
getpos(son[u],sp);
for(int i = head[u]; ~i ; i = edge[i].next)
{
int v = edge[i].to;
if(v != son[u] && v != fa[u])
getpos(v,v);
}
}
struct node
{
int l,r,mid;
ll Max,Min;
ll lans,rans;
ll add;
} tree[maxn << 2];
void push_up(int i)
{
tree[i].Max = MAX(tree[lson].Max,tree[rson].Max);
tree[i].Min = MIN(tree[lson].Min,tree[rson].Min);
tree[i].rans = MAX(tree[rson].Max - tree[lson].Min,MAX(tree[lson].rans,tree[rson].rans));
tree[i].lans = MAX(tree[lson].Max - tree[rson].Min,MAX(tree[lson].lans,tree[rson].lans));
if(tree[i].lans < 0) tree[i].lans = 0;
if(tree[i].rans < 0) tree[i].rans = 0;
}
void build(int i,int l,int r)
{
tree[i].l = l,tree[i].r = r;
tree[i].mid=(l+r) >>1;
tree[i].add = 0;
tree[i].Max = 0,tree[i].Min = inf;
tree[i].lans = 0,tree[i].rans = 0;
if(l == r)
{
tree[i].Max = tree[i].Min = a[fp[l]];
return;
}
build(lson,l,tree[i].mid);
build(rson,tree[i].mid+1,r);
push_up(i);
}
void update(int i,ll d)
{
tree[i].Max += d,tree[i].Min += d;
tree[i].add += d;
}
void push_down(int i)
{
if(tree[i].add)
{
update(lson,tree[i].add),update(rson,tree[i].add);
tree[i].add = 0;
}
}
void update_area(int i,int l,int r,ll val)
{
if(tree[i].l >= l && tree[i].r <= r)
{
update(i,val);
return ;
}
push_down(i);
int mid = tree[i].mid;
if(l <= mid)
update_area(lson,l,r,val);
if(r > mid)
update_area(rson,l,r,val);
push_up(i);
}
ll query(int i,int l,int r,int flag,ll& MaxPrice,ll& MinPrice)
{
if(tree[i].l >= l && tree[i].r <= r)
{
MinPrice = tree[i].Min;
MaxPrice = tree[i].Max;
if(flag)
{
return tree[i].rans;
}
else
{
return tree[i].lans;
}
}
push_down(i);
int mid = tree[i].mid;
if(r <= mid)
return MAX(0LL,query(lson,l,r,flag,MaxPrice,MinPrice));
else if(l > mid)
return MAX(0LL,query(rson,l,r,flag,MaxPrice,MinPrice));
else
{
ll ta = 0;
ll max1,max2,min1,min2;
ll ans = MAX(query(lson,l,mid,flag,max1,min1),query(rson,mid+1,r,flag,max2,min2));
if(flag)
ta = max2 - min1;
else
ta = max1 - min2;
MaxPrice = MAX(max1,max2);
MinPrice = MIN(min1,min2);
ta = MAX(ta,0LL);
return MAX(ans,ta);
}
push_up(i);
}
void make_add(int u,int v,ll val)
{
int f1 = top[u],f2 = top[v];
while(f1 != f2)
{
if(dep[f1] < dep[f2])
{
swap(f1,f2),swap(u,v);
}
update_area(1,p[f1],p[u],val);
u = fa[f1],f1 = top[u];
}
if(dep[u] > dep[v])
swap(u,v);
update_area(1,p[u],p[v],val);
return ;
}
ll make_query(int u,int v)
{
ll tmin,tmax,tMin,tMax;
ll cmin,cmax,cMin,cMax;
tMin = tmin = tree[1].Max;
tMax = tmax = 0;
ll cnt = 0;
int f1 = top[u],f2 = top[v];
while(f1 != f2)
{
if(dep[f1] > dep[f2])
{
// cout << p[f1] <<" "<<p[u] <<endl;
cnt = MAX(cnt,query(1,p[f1],p[u],0,cmax,cmin));
cnt = MAX(cnt,cmax - tmin);
cnt = MAX(cnt,tMax - cmin);
tmin = MIN(cmin,tmin);
tmax = MAX(cmax,tmax);
u = fa[f1],f1 = top[u];
// tmax = max(tmax,cmax);
}
else
{
// cout << p[f2] <<" "<<p[v] <<endl;
cnt = MAX(cnt,query(1,p[f2],p[v],1,cMax,cMin));
cnt = MAX(cnt,tMax - cMin);
cnt = MAX(cnt,cMax-tmin);
tMax = MAX(tMax,cMax);
tMin = MIN(tMin,cMin);
v = fa[f2],f2 = top[v];
// tMin = min(tMin,cMin);
}
}
if(dep[u] > dep[v])
{
cnt =MAX(cnt,query(1,p[v],p[u],0,cmax,cmin));
cnt =MAX(cnt,cmax-tmin);
tmin = MIN(tmin,cmin);
cnt = MAX(cnt,tMax-tmin);
}
else
{
cnt =MAX(cnt,query(1,p[u],p[v],1,cMax,cMin));
// cout <<"max" <<cMax <<" " <<"min" <<cMin <<endl;
cnt = MAX(cnt,tMax-cMin);
tMax = MAX(tMax,cMax);
cnt = MAX(cnt,tMax-tmin);
}
return cnt;
}
/*
5
3 1 1 1
1 2 2 3
3
1 1 500000000
2 1 1
3 1 1
*/
int main()
{
// freopen("in.txt","r",stdin);
int T;
int m,u,v;
ll w;
scanf("%d",&T);
while(T--)
{
ini();
scanf("%d",&n);
for(int i = 1; i <= n; i++)
scanf("%I64d",&a[i]);
for(int i =1; i <n; i++)
{
scanf("%d%d",&u,&v);
add_edge(u,v);
add_edge(v,u);
}
dfs1(1,0,0);
getpos(1,1);
build(1,1,pos-1);
scanf("%d",&m);
for(int i = 1; i <= m; i++)
{
scanf("%d%d%I64d",&u,&v,&w);
printf("%I64d\n",make_query(u,v));
make_add(u,v,w);
}
}
return 0;
}
