FZU Problem 2082 过路费 树链剖分

Problem 2082 过路费

 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。

 Output

对于每个询问，输出一行，表示最少要花的过路费。

 Sample Input

2 3 1 2 1 1 1 2 0 1 2 1 2 1

 Sample Output

1 2

 Source

FOJ有奖月赛-2012年4月（校赛热身赛）

题解

　　裸题

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;

#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair

typedef long long LL;
const long long INF = 1e18;
const double Pi = acos(-1.0);
const int N = 2e5+, M = 5e5+, inf = 2e9, mod = ;

struct edge{int to,next,value;}e[N * ];

struct Lin {int u; int v; int w;
    Lin(int u = , int v = , int w = ) : u(u), v(v), w(w) {}
}L[N*];
LL sum[N];
int head[N],t=,f[N],q,n,top[N],siz[N],son[N],pos[N],val[N],deep[N],tot;
void init() {
    t = ;
    memset(head,,sizeof(head));
    deep[] = ;
    siz[] = ;
    f[] = ;
    tot = ;
}
void add(int u,int v)
{e[t].next=head[u];e[t].to=v;head[u]=t++;}

void dfs1(int u,int fa) {
        siz[u] = ;
        son[u] = ;
        f[u] = fa;
        deep[u] = deep[fa] + ;
        for(int i = head[u]; i; i = e[i].next) {
            int to = e[i].to;
            if(to == fa) continue;
            dfs1(to,u);
            siz[u] += siz[to];
            if(siz[to] > siz[son[u]]) son[u] = to;
        }
}
void dfs2(int u,int chan) {
        top[u] = son[chan]==u?top[chan]:u;
        pos[u] = ++tot;
        if(son[u]) dfs2(son[u],u);
        for(int i = head[u]; i; i = e[i].next) {
            int to = e[i].to;
            if(to == son[u] || to == chan) continue;
            dfs2(to,u);
        }
}
void build(int i,int ll,int rr)
{
    sum[i] = ;
    if(ll == rr) {
        sum[i] = val[ll];
        return ;
    }
    build(ls,ll,mid), build(rs,mid+,rr);
    sum[i] = sum[ls] + sum[rs];
}
void update(int i,int ll,int rr,int x,int c) {
    if(x == ll && x == rr) {
        sum[i] = c;
        return ;
    }
    if(x <= mid) update(ls,ll,mid,x,c);
    else update(rs,mid+,rr,x,c);
    sum[i] = sum[ls] + sum[rs];
}
LL query(int i,int ll,int rr,int x,int y) {
    if(x == ll && rr == y) {
        return sum[i];
    }
    if(y <= mid) return query(ls,ll,mid,x,y);
    else if(x > mid) return query(rs,mid+,rr,x,y);
    else return query(ls,ll,mid,x,mid) + query(rs,mid+,rr,mid+,y);
}
LL sub_query(int x,int y,LL ret = ) {
        while (top[x] != top[y])
        {
            if(deep[top[x]] < deep[top[y]]) swap(x,y);
            ret += query(,,n,pos[top[x]],pos[x]);
            x = f[top[x]];
        }
        if(x == y) return ret;
        if(deep[x] > deep[y]) swap(x,y);
        return ret + query(,,n,pos[x]+, pos[y]);
}
void solve() {
        init();
        for(int i = ; i <= n-; ++i) {
                int a,b,c;
                scanf("%d%d%d",&a,&b,&c);
                L[i] = Lin(a,b,c);
                add(a,b);add(b,a);
        }
        dfs1(,);
        dfs2(,);
        val[] = ;
        for(int i = ; i <= n-; ++i) {
            if(deep[L[i].u] < deep[L[i].v]) swap(L[i].u, L[i].v);
            val[pos[L[i].u]] = L[i].w;
        }//cout<<1<<endl;
        build(,,n);

        while(q--){
            int op,x,y;
            scanf("%d%d%d",&op,&x,&y);
            if(op == )
                update(,,n,pos[L[x].u],y);
            else printf("%I64d\n",sub_query(x,y));
        }
}
int main() {
        while(~scanf("%d%d",&n,&q)) {solve();}return ;
}