模板【HDU5893 2016 ACM ICPC Asia Regional Shenyang Online B】【树链剖分 权在边上 链值合并 结构体写法】List wants to travel

List wants to travel

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 727    Accepted Submission(s): 179


[align=left]Problem Description[/align]
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.)
 

[align=left]Input[/align]
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).
 

[align=left]Output[/align]
He insure you that there is exactly one route between every two different cities.
 

[align=left]Sample Input[/align]

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

 

[align=left]Sample Output[/align]

3
2

 

[align=left]Source[/align]
2016 ACM/ICPC
Asia Regional Shenyang Online

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<ctype.h>
#include<math.h>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<string>
#include<algorithm>
#include<time.h>
#include<bitset>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x,y) memset(x,y,szof(x))
#define MC(x,y) memcpy(x,y,szof(x))
#define MP(x,y) make_pair(x,y)
#define ls o<<1
#define rs o<<1|1
#define lson o<<1,l,mid
#define rson o<<1|1,mid+1,r
#define ff first
#define ss second
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T> inline void gmax(T &a, T b) { if (b>a)a = b; }
template <class T> inline void gmin(T &a, T b) { if (b<a)a = b; }
const int N = 1e5 + 10, M = N * 2, Z = 1e9 + 7, maxint = 2147483647, ms31 = 522133279, ms63 = 1061109567, ms127 = 2139062143; const double eps = 1e-8, PI = acos(-1.0);//.0
int casenum, casei;
int n, m, id, tim;
int x, y, z; char s[10];
int first
, w[M], c[M], nxt[M];
int u
, v
, fa
, dep
, sz
, pos
, top
;
void ins(int x, int y, int z)
{
id++;
w[id] = y;
c[id] = z;
nxt[id] = first[x];
first[x] = id;
}

//u[x]表示边的编号为x，对应的节点是几号
void dfs1(int x)
{
sz[x] = 1;
for (int z = first[x]; z; z = nxt[z])
{
int y = w[z];
if (y == fa[x])continue;
fa[y] = x;
dep[y] = dep[x] + 1;
u[z] = u[z ^ 1] = y;
dfs1(y);
sz[x] += sz[y];
}
}
void dfs2(int x, int chain)
{
pos[x] = ++tim;
top[x] = chain;
int p = 0;
for (int z = first[x]; z; z = nxt[z])
{
int y = w[z];
if (y != fa[x] && sz[y]>sz[p])p = y;
}
if (p == 0)return;
dfs2(p, chain);
for (int z = first[x]; z; z = nxt[z])
{
int y = w[z];
if (y != fa[x] && y != p)dfs2(y, y);
}
}
int flag[1 << 18];
struct node
{
int num, lc, rc;
}a[1 << 18];
void pushdown(int o)
{
if (flag[o] != 1e9)
{
flag[ls] = flag[rs] = flag[o];
a[ls] = a[rs] = { 1,flag[o],flag[o] };
flag[o] = 1e9;
}
}
node merge(node lc, node rc)
{
return{ lc.num + rc.num - (lc.rc == rc.lc), lc.lc, rc.rc };
}
void build(int o, int l, int r)
{
flag[o] = 1e9;
if (l == r)
{
a[o] = { 1,v[l],v[l] };
return;
}
int mid = (l + r) >> 1;
build(lson);
build(rson);
a[o] = merge(a[ls], a[rs]);
}
int L, R, V;
void color(int o, int l, int r)
{
if (l >= L && r <= R)
{
a[o] = { 1,V,V };
flag[o] = V;
return;
}
pushdown(o);
int mid = (l + r) >> 1;
if (L <= mid)color(lson);
if (R > mid)color(rson);
a[o] = merge(a[ls], a[rs]);
}
node check(int o, int l, int r)
{
if (l >= L && r <= R)return a[o];
pushdown(o);
int mid = (l + r) >> 1;
if (R <= mid)return check(lson);
else if (L > mid)return check(rson);
else return merge(check(lson), check(rson));
}
void COLOR(int x, int y, int z)
{
while (top[x] != top[y])
{
if (dep[top[x]]<dep[top[y]])swap(x, y);
L = pos[top[x]]; R = pos[x]; V = z; color(1, 1, n);
x = fa[top[x]];
}
if (pos[x]>pos[y])swap(x, y);
if (pos[x] < pos[y])
{
L = pos[x] + 1; R = pos[y]; V = z; color(1, 1, n);
}
}
vector< pair<int, int> >up;
vector< pair<int, int> >dn;
int CHECK(int x, int y)
{
if (x == y)return 0;
up.clear();
dn.clear();
while (top[x] != top[y])
{
if (dep[top[x]] > dep[top[y]])
{
up.push_back(MP(pos[top[x]], pos[x]));
x = fa[top[x]];
}
else
{
dn.push_back(MP(pos[top[y]], pos[y]));
y = fa[top[y]];
}
}
if (pos[x] < pos[y])dn.push_back(MP(pos[x] + 1, pos[y]));
else if (pos[y] < pos[x])up.push_back(MP(pos[y] + 1, pos[x]));
node lft = { 0,-1,-1 };
node rgt = { 0,-2,-2 };
for (int i = 0, g = up.size(); i < g; ++i)
{
tie(L, R) = up[i];
lft = merge(check(1, 1, n), lft);
}
for (int i = 0, g = dn.size(); i < g; ++i)
{
tie(L, R) = dn[i];
rgt = merge(check(1, 1, n), rgt);
}
swap(lft.lc, lft.rc);
return merge(lft, rgt).num;
}
int main()
{
//fre();
while (~scanf("%d%d", &n, &m))
{
memset(first, 0, (n + 2) * 4); id = 1;
for (int i = 1; i<n; i++)
{
scanf("%d%d%d", &x, &y, &z);
ins(x, y, z);
ins(y, x, z);
}
fa[1] = 0; dep[1] = 0; dfs1(1);
tim = 0; dfs2(1, 1);
//u[i<<1]是节点
//pos[节点]是节点对应在线段树上的位置
//v[位置]对应的是权值
v[1] = 1e9; for (int i = 1; i < n; i++)v[pos[u[i << 1]]] = c[i << 1];
build(1, 1, n);
while (m--)
{
scanf("%s", s);
if (s[0] == 'C')scanf("%d%d%d", &x, &y, &z), COLOR(x, y, z);
else scanf("%d%d", &x, &y), printf("%d\n", CHECK(x, y));
}
}
return 0;
}
/*
【trick&&吐槽】
链的合并，还是取出lca，然后两端链合并得好

【题意】
给你一棵树，有n(1e5)个节点和m(1e5)个操作
操作1，链（x，y）染色为z([0,1e9])
操作2，问你链（x，y）的颜色段数

【类型】
树链剖分

【分析】
树链剖分+求区间段数，记录区间左右颜色即可。

【时间复杂度&&优化】
O(mlognlogn)

*/