FZU 2082 树链剖分 区间最大和

Problem Description

有n座城市，由n-1条路相连通，使得任意两座城市之间可达。每条路有过路费，要交过路费才能通过。每条路的过路费经常会更新，现问你，当前情况下，从城市a到城市b最少要花多少过路费。

 Input

有多组样例，每组样例第一行输入两个正整数n,m(2 <= n<=50000，1<=m <= 50000),接下来n-1行，每行3个正整数a b c，(1 <= a,b <= n , a != b , 1 <= c <= 1000000000).数据保证给的路使得任意两座城市互相可达。接下来输入m行，表示m个操作，操作有两种：一. 0 a b，表示更新第a条路的过路费为b，1 <= a <= n-1 ； 二. 1 a b ， 表示询问a到b最少要花多少过路费。

树链剖分..

＝＝debug了好久。。就是自己逗。。 QUERY函数写错了。。

#include <cstdio>#include <cstdlib>#include <cstring>#include <vector>#include <iostream>using namespace std;const int N = 155555;const int M = 1111111;struct Edge{int u,v,w;Edge(){}Edge(int _l,int _r,int _w){u=_l, v=_r,w=_w;}}edge[M];struct node{int to,w;node(){}node(int _to,int _w){to=_to,w=_w;}};vector<node> V[N];int n,m;int dep[N],fa[N],sz[N],son[N];int tim[N],ra[N],top[N];int tx;int val[N];void dfs1(int x,int Fa,int Deep){fa[x] = Fa , sz[x] = 1 , dep[x] = Deep;for(int i = 0;i < V[x].size();i++){int v = V[x][i].to;if( v!=Fa){val[v] = V[x][i].w;dfs1(v,x,Deep+1);sz[x] += sz[v];if(son[x]==-1 || sz[v] > sz[son[x]]){son[x] = v;}}}}void dfs2(int u,int Top){tim[u] = ++tx; ra[tx] = u;top[u] = Top;if(son[u]!=-1)dfs2(son[u],Top);for(int i = 0;i < V[u].size();i++){int v = V[u][i].to;if(v!=fa[u] && v!=son[u]){dfs2(v,v);}}}#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define ll long longll sum[N<<2];void Push_up(int rt){sum[rt] = sum[rt<<1] + sum[rt<<1|1];}void build(int l,int r,int rt){if(l==r){sum[rt] = val[ra[l]];return ;}int m=(l+r)>>1;build(lson);build(rson);Push_up(rt);}int ql,qr;int qv;void update(int l,int r,int rt){if(l==r){sum[rt] = qv;return;}int m = (l+r)>>1;if(ql<=m)update(lson);if(ql>m)update(rson);Push_up(rt);}ll query(int l,int r,int rt){if(ql<=l && r <= qr){return sum[rt];}int m=(l+r)>>1;ll ans = 0;if(ql<=m) ans+=query(lson);if(qr>m) ans+=query(rson);return ans;}void UPDATE(int idx,int w){int u = edge[idx].u, v = edge[idx].v;if(dep[u] < dep[v]) swap(u,v);val[u] = w;qv = w;ql=tim[u];update(1,n,1);}ll QUERY(int x,int y){int f1 = top[x] , f2 = top[y];long long ans = 0;while(f1!=f2){if(dep[f1]<dep[f2]) swap(f1,f2),swap(x,y);ql = tim[f1] , qr = tim[x];ans+=query(1,n,1);x = fa[f1];f1 = top[x];}if(dep[x] > dep[y]) swap(x,y);ql = tim[x], qr = tim[y] , ans += query(1,n,1);ans -= val[x];return ans;}int main(){while(scanf("%d %d",&n,&m)!=EOF){for(int i=0;i<=n;i++) V[i].clear();for(int i=1,u,v,w;i<n;i++){scanf("%d %d %d",&u,&v,&w);edge[i] = Edge(u,v,w);V[u].push_back(node(v,w));V[v].push_back(node(u,w));}memset(son,-1,sizeof(son));tx = 0;dfs1(1,0,1);dfs2(1,1);val[1]=0;build(1,n,1);for(int qq,u,w; m--; ){scanf("%d %d %d",&qq,&u,&w);if(qq==0){UPDATE(u,w);}else{cout << QUERY(u,w) << endl;}}}return 0;}