Hdu-5893 List wants to travel(树链剖分)

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 31 2 22 3 11 7 21 4 23 5 23 6 15 8 25 9 3Query 1 8Change 2 6 3Query 1 6

 

Sample Output

32

 

Source

2016 ACM/ICPC Asia Regional Shenyang Online

题意：给一棵边权数，q个操作，每次询问(u,v)路径上有多少段不同边权，或者修改(u,v)上的所有边权。


分析：比赛时从头WA到尾，学树剖时只是草草地熟悉了一下模板，没有深入理解它的原理，所以改这道题时出现了很大的问题，感觉自己还是太浮躁了。

#include<iostream>#include<string>#include<algorithm>#include<cstdlib>#include<cstdio>#include<set>#include<map>#include<vector>#include<cstring>#include<stack>#include<cmath>#include<queue>#define INF 0x3f3f3f3f#define N 40005using namespace std;int dep[N],siz[N],fa[N],id[N],son[N],val[N],top[N],fid[N];//深度，儿子数量，父亲，新编号，重儿子，权值，所属链头，反编号。 int T,n,q,num;vector <int> G[N];struct Edge{int x,y,val;} edge[N];struct Segtree{int l,r,down,lc,rc,ans;bool same;} tree[4*N];void dfs1(int u,int f,int d){dep[u] = d;siz[u] = 1;son[u] = 0;fa[u] = f;for(int v : G[u]) if(v != f) {  dfs1(v,u,d+1);  siz[u] += siz[v];  if(siz[son[u]] < siz[v]) son[u] = v; }}void dfs2(int u,int tp){top[u] = tp;id[u] = ++num;fid[id[u]] = u;if(son[u]) dfs2(son[u],tp);for(int v : G[u]) if(v != fa[u] && v != son[u]) dfs2(v,v);}void push_up(int i){tree[i].lc = tree[2*i].lc;tree[i].rc = tree[2*i+1].rc;tree[i].ans = tree[2*i].ans + tree[2*i+1].ans - (tree[2*i].rc == tree[2*i+1].lc);tree[i].same = tree[2*i].same && tree[2*i+1].same && (tree[2*i].lc == tree[2*i+1].rc);}void build(int i,int l,int r){tree[i].l = l;tree[i].r = r;tree[i].down = 0;if(l == r) {tree[i].ans = 1;tree[i].lc = tree[i].rc = val[fid[l]];tree[i].same = true;return;}int mid=(l+r)>>1;build(2*i,l,mid);build(2*i+1,mid+1,r);push_up(i);}void Down(int u){if(!tree[u].down) return;tree[2*u].down = tree[2*u+1].down = tree[u].down;tree[2*u].ans = tree[2*u+1].ans = 1;tree[2*u].lc = tree[2*u].rc = tree[2*u+1].lc = tree[2*u+1].rc = tree[u].down;tree[2*u].same = tree[2*u+1].same = true;tree[u].down = 0;}int query(int i,int x,int y){int l = tree[i].l,r = tree[i].r;if(l == x && r == y) return tree[i].ans;Down(i);int mid = (l+r)/2;if(y <= mid) return query(2*i,x,y);else  if(x <= mid) return query(2*i,x,mid) + query(2*i+1,mid+1,y) - (tree[2*i].rc == tree[2*i+1].lc); else return query(2*i+1,x,y);}int query_point(int i,int x){int l = tree[i].l,r = tree[i].r;if(tree[i].same) return tree[i].lc;Down(i);int mid = (l+r)/2;if(x <= mid) return query_point(2*i,x);else return query_point(2*i+1,x);}void deal(int i,int x,int y,int val){int l = tree[i].l,r = tree[i].r;if(l == x && r == y){tree[i].down = val;tree[i].same = true;tree[i].ans = 1;tree[i].lc = tree[i].rc = val;return;}Down(i);int mid = (l+r)/2;if(y <= mid) deal(2*i,x,y,val);else  if(x <= mid) { deal(2*i,x,mid,val);deal(2*i+1,mid+1,y,val); } else deal(2*i+1,x,y,val);push_up(i);}int Yougth_ask(int u,int v){int tp1 = top[u],tp2 = top[v];int ans = 0,pre1 = 0,pre2 = 0; while(tp1 != tp2){if(dep[tp1] < dep[tp2]){swap(tp1,tp2);swap(u,v);swap(pre1,pre2);}ans += query(1,id[tp1],id[u]);if(pre1) ans -= (query_point(1,id[u]) == pre1);pre1 = query_point(1,id[tp1]);u = fa[tp1];tp1 = top[u]; }if(u == v) return (ans - (pre1 == pre2));if(dep[u] > dep[v]) {swap(u,v);swap(pre1,pre2);}ans += query(1,id[son[u]],id[v]);if(pre2) ans -= (query_point(1,id[v]) == pre2);if(pre1) ans -= (query_point(1,id[son[u]]) == pre1);return ans;}void Yougth_deal(int u,int v,int val){int tp1 = top[u],tp2 = top[v];while(tp1 != tp2){if(dep[tp1] < dep[tp2]){swap(tp1,tp2);swap(u,v);}deal(1,id[tp1],id[u],val);u = fa[tp1];tp1 = top[u]; }if(u == v) return;if(dep[u] > dep[v]) swap(u,v);deal(1,id[son[u]],id[v],val);}int main(){while(~scanf("%d%d",&n,&q)){for(int i = 1;i <= n;i++) G[i].clear();for(int i = 1;i < n;i++){scanf("%d%d%d",&edge[i].x,&edge[i].y,&edge[i].val);edge[i].val++;G[edge[i].x].push_back(edge[i].y);G[edge[i].y].push_back(edge[i].x);}num = 0;dfs1(1,0,1);dfs2(1,1);for(int i = 1;i < n;i++) if(dep[edge[i].x] > dep[edge[i].y]) val[edge[i].x] = edge[i].val; else val[edge[i].y] = edge[i].val;build(1,1,num);char s[20];for(int i = 1;i <= q;i++){scanf("%s",s);if(s[0] == 'Q'){ int x,y;scanf("%d%d",&x,&y);printf("%d\n",Yougth_ask(x,y));}else {int x,y,val;scanf("%d%d%d",&x,&y,&val);val++;Yougth_deal(x,y,val);}}}}