List wants to travel

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 429    Accepted Submission(s): 92

Problem Description
A boy named List who is perfect in English. Now he wants to travel and he is making a plan. But the cost of living in same citie always changes. Now he wants to know how many different kinds of continuous same cost he has to pay for living between two cities. Can you help him? (He is so lazy to do this by himself.)
 
Input
There are multiple cases. The first line contains two positive numbers N and M(N (N<=40000) where N is the amount of cities and M (M<=50000)) is the amount of operations.Then N-1 lines where each line have 3 integers a b and c, representing that there is a bidirectionoal road between city a and city b, and the cost is c.(a != b and c <= 100000). Then there are M lines of operation. For example, "Change a b c" means changing all the costs of the road which are passed by him when he travels from city a to city b to c. "Query a b" means he wants you to tell him how many different kinds of continuous same cost he has to pay traveling from city a to city b.(if a == b, the cost is 0).
 
Output
He insure you that there is exactly one route between every two different cities.
 
Sample Input
9 3
1 2 2
2 3 1
1 7 2
1 4 2
3 5 2
3 6 1
5 8 2
5 9 3
Query 1 8
Change 2 6 3
Query 1 6
 
Sample Output
3
2
 
Source
/*
hdu 5893 (树链剖分+合并) problem:
1.update:将a->b的边权设置为c
2.query:求a->b的连续边权的个数
222333 -> 2 22112->3 solve:
主要是查询的意思不是很懂. how many different kinds of continuous same cost
以为不同还要分长度,数值大小什么的。 于是没怎么想
结果后来发现是求区间中有多少个连续的子区间 - -. 感觉很僵
update直接用一个标记解决
query的时候, 维护区间左右端点的值以合并区间,合并的时候注意维护子区间的数量
而且树链剖分时 在合并链的时候也要进行判断什么的.
update的u == v的时候最好判断一下, 否则查询son[u] -> v会有问题
我们只有一个重链,可能有很多的轻链,query的 u == v时就是轻链合并的情况.
总体都是线段树的思路 hhh-2016-09-19 22:36:11
*/
#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]
using namespace std;
const int maxn = 200100;
const int inf = 0x3f3f3f3f;
int head[maxn],tot,pos,son[maxn];
int top[maxn],fp[maxn],fa[maxn],dep[maxn],num[maxn],p[maxn];
int n;
int a[maxn];
struct Edge
{
int to,next;
int w;
} edge[maxn<<2]; void ini()
{
tot = 0,pos = 1;
clr(head,-1),clr(son,-1);
clr(a,0);
} void add_edge(int u,int v,int w)
{
edge[tot].to = v,edge[tot].next = head[u],edge[tot].w = w,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)
{
a[v] = edge[i].w;
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;
int ls,rs,same;
ll num ;
} tree[maxn << 2]; void push_up(int i)
{ if(tree[lson].rs == tree[rson].ls )
{
tree[i].num = tree[lson].num + tree[rson].num -1;
}
else
tree[i].num =tree[lson].num + tree[rson].num;
tree[i].ls = tree[lson].ls;
tree[i].rs = tree[rson].rs;
} void build(int i,int l,int r)
{
tree[i].l = l,tree[i].r = r;
tree[i].mid=(l+r) >>1;
tree[i].same = inf;
tree[i].num = tree[i].ls = tree[i].rs = 0;
if(l == r)
{
tree[i].ls = tree[i].rs = a[fp[l]];
tree[i].num = 1;
// cout << fp[l] <<" " << a[fp[l]] <<endl;
return;
}
build(lson,l,tree[i].mid);
build(rson,tree[i].mid+1,r);
push_up(i);
}
void make_same(int i,int val)
{
tree[i].same = val;
tree[i].num = 1;
tree[i].ls = tree[i].rs = val;
}
void push_down(int i)
{
if(tree[i].same != inf)
{
make_same(lson,tree[i].same);
make_same(rson,tree[i].same);
tree[i].same = inf;
}
} void update_area(int i,int l,int r,int val)
{
if(tree[i].l >= l && tree[i].r <= r)
{
tree[i].same = val;
tree[i].num = 1;
tree[i].ls = tree[i].rs = 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 &tls,int &trs)
{
if(tree[i].l >= l && tree[i].r <= r)
{
if(tree[i].l == l)
tls = tree[i].ls;
if(tree[i].r == r)
trs = tree[i].rs;
return tree[i].num ;
}
push_down(i);
int mid = tree[i].mid ;
ll ans = 0;
if(r <= mid)
ans = query(lson,l,r,tls,trs);
else if(l > mid)
ans = query(rson,l,r,tls,trs);
else
{
int tls1,tls2,trs1,trs2;
ll t1 = query(lson,l,mid,tls1,trs1);
ll t2 = query(rson,mid+1,r,tls2,trs2);
ans = t1 + t2;
if(tree[lson].rs == tree[rson].ls)
{
ans --;
}
tls = tls1,trs = trs2;
}
push_up(i);
return ans;
} void update_same(int u,int v,int 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(u == v)
return ;
if(dep[u] > dep[v]) swap(u,v);
update_area(1,p[son[u]],p[v],val);
} ll query_dif(int u,int v)
{
int f1 = top[u],f2 = top[v];
int ls1,rs1,ls2,rs2;
ls1 = rs1 = ls2 = rs2 = inf;
int tls1,trs1,tls2,trs2;
int fi1 = 1,fi2 = 1;
ll ans = 0;
while(f1 != f2)
{
if(dep[f1] > dep[f2])
{
ans += query(1,p[f1],p[u],tls1,trs1);
if(trs1 == ls1)
ans--;
ls1 = tls1;
if(fi1)
{
rs1 = trs1;
fi1 = 0;
}
u = fa[f1],f1 = top[u];
}
else
{
ans += query(1,p[f2],p[v],tls2,trs2);
if(trs2 == ls2)
ans--;
ls2 = tls2;
if(fi2)
{
rs2 = trs2;
fi2 = 0;
}
v = fa[f2],f2 = top[v];
}
}
if(u == v)
{
if(tls1 == tls2)
ans --;
return ans;
}
if(dep[u] > dep[v])
{
ans += query(1,p[son[v]],p[u],tls1,trs1);
if(trs1 == ls1)
ans--;
if(tls1 == ls2)
ans --;
}
else
{
ans += query(1,p[son[u]],p[v],tls2,trs2);
if(trs2 == ls2)
ans--;
if(tls2 == ls1)
ans--;
}
return ans;
}
char str[10];
int main()
{
// freopen("in.txt","r",stdin);
int a,b,c;
int m,u,v,w;
while(scanf("%d%d",&n,&m) != EOF)
{
ini();
for(int i =1; i <n; i++)
{
scanf("%d%d%d",&u,&v,&w);
add_edge(u,v,w);
add_edge(v,u,w);
}
dfs1(1,0,0);
getpos(1,1);
build(1,1,pos-1);
// cout << pos -1 <<endl;
for(int i = 1; i <= m; i++)
{
scanf("%s",str);
scanf("%d%d",&a,&b);
if(str[0] == 'C')
{
scanf("%d",&c);
update_same(a,b,c);
}
else
{
if(a == b)
printf("0\n");
else
printf("%I64d\n",query_dif(a,b));
}
}
}
return 0;
}

  

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