Codeforces 838B | 易学教程

题目链接：http://codeforces.com/problemset/problem/838/B

You are given a directed weighted graph with n nodes and 2n - 2 edges. The nodes are labeled from 1 to n, while the edges are labeled from 1 to 2n - 2. The graph's edges can be split into two parts.

The first n - 1 edges will form a rooted spanning tree, with node 1 as the root. All these edges will point away from the root.
The last n - 1 edges will be from node i to node 1, for all 2 ≤ i ≤ n.

You are given q queries. There are two types of queries

1 i w: Change the weight of the i-th edge to w
2 u v: Print the length of the shortest path between nodes u to v

Given these queries, print the shortest path lengths.

Input

The first line of input will contain two integers n, q (2 ≤ n, q ≤ 200 000), the number of nodes, and the number of queries, respectively.

The next 2n - 2 integers will contain 3 integers a_i, b_i, c_i, denoting a directed edge from node a_i to node b_i with weight c_i.

The first n - 1 of these lines will describe a rooted spanning tree pointing away from node 1, while the last n - 1 of these lines will have b_i = 1.

More specifically,

The edges (a₁, b₁), (a₂, b₂), ... (a_n - 1, b_n - 1) will describe a rooted spanning tree pointing away from node 1.
b_j = 1 for n ≤ j ≤ 2n - 2.
a_n, a_n + 1, ..., a_2n - 2 will be distinct and between 2 and n.

The next q lines will contain 3 integers, describing a query in the format described in the statement.

All edge weights will be between 1 and 10⁶.

Output

For each type 2 query, print the length of the shortest path in its own line.

Example

Input

Output

题意：

给出n个节点，2n-2条边（有向带权）；

其中前n-1条边使得n个点构成一棵带权有向树；后n-1条边，权重为w，是从2~n号节点直接连接向1号节点的；

现在给出两种操作：

①修改第 i 条边的权重为w；

②查询节点u和v之间最短路径的权重和；

题解：

先忽略后n-1条边，DFS序拍平整棵树；

同时在DFS时，顺便计算出节点i的dist[i]，代表从节点1到节点i的唯一的一条路径的权重和；

记录每个节点的 dist[i] + Edge(i→1).weight，用线段树在DFS序上维护区间最小值；

then……

对于修改操作：

　　①若修改的边是Edge(x→1)类型的，那么更新区间[ in[x] , in[x] ]上的值；

　　②否则，更新区间[ in[x] , out[x] ]上的值；

对于查询操作：

　　①v在u统领的子树内，按道理来讲直接dist[v]-dist[u]即可，

　　　但是我们在修改边权时不对dist[]数组进行更新，所以要通过式子dist[x] = query([ in[x] , in[x] ]) - Edge(x→1).weight来求得dist[u]和dist[v]；

　　②v不在u统领的子树内，那么只能通过 u → … → 1 → … → v 这样的路径从u走到v，

　　　显然，u统领的子树内的每个节点，都能使得从节点u出发回到节点1，

　　　那么我们就query([ in[u] , out[u] ])查：u统领的这颗子树内的哪个节点，它的 dist[x] + Edge(x→1).weight 是最小的；

　　　一旦查到，ans = query([ in[u] , out[u] ]) - dist[u] + dist[v]，同样的道理，这里的dist[u]和dist[v]都要像上面那样通过式子 dist[x] = query([ in[x] , in[x] ]) - Edge(x→1).weight 求得。

AC代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn=200000+10;
const LL INF=1e18;

int n,q;

//邻接表
struct Edge{
    int u,v;
    LL w;
    Edge(int u,int v,LL w)
    {
        this->u=u;
        this->v=v;
        this->w=w;
    }
};
vector<Edge> E; int E_size;
vector<int> G[maxn];
int ptr[2*maxn];
void adjListInit(int l,int r)
{
    E.clear(); E_size=0;
    for(int i=l;i<=r;i++) G[i].clear();
}
void addEdge(int u,int v,LL w,int i)
{
    E.push_back(Edge(u,v,w)); E_size++;
    ptr[i]=E_size-1;
    G[u].push_back(E_size-1);
}

//存储所有返回到1的边（即编号为n~2n-2的边）
vector<Edge> Eback;
int Gback[maxn];
int Eback_size;

//DFS建立DFS序列，以及计算深度
LL dist[maxn];
int in[maxn],out[maxn];
int peg[maxn];
int dfs_clock;
inline void dfsInit()
{
    dist[1]=0;
    dfs_clock=0;
}
void dfs(int now,int par)
{
    //printf("now=%d\n",now);
    in[now]=++dfs_clock;
    peg[in[now]]=now;

    for(int i=0,_size=G[now].size();i<_size;i++)
    {
        Edge &e=E[G[now][i]]; int nxt=e.v;
        if(nxt!=par)
        {
            dist[nxt]=dist[now]+e.w;
            dfs(nxt,now);
        }
    }

    out[now]=dfs_clock;
}

//线段树
struct Node{
    int l,r;
    LL val,lazy;
    void update(LL x)
    {
        val+=x;
        lazy+=x;
    }
}node[4*maxn];
void pushdown(int root)
{
    if(node[root].lazy)
    {
        node[root*2].update(node[root].lazy);
        node[root*2+1].update(node[root].lazy);
        node[root].lazy=0;
    }
}
void pushup(int root)
{
    node[root].val=min(node[root*2].val,node[root*2+1].val);
}
void build(int root,int l,int r)
{
    node[root].l=l; node[root].r=r;
    node[root].val=0; node[root].lazy=0;
    if(l==r)
    {
        int p=peg[l]; //从DFS序中的位置逆向查节点编号
        node[root].val=dist[p]+Eback[Gback[p]].w;
    }
    else
    {
        int mid=l+(r-l)/2;
        build(root*2,l,mid);
        build(root*2+1,mid+1,r);
        pushup(root);
    }
}
void update(int root,int st,int ed,int val)
{
    if(st>node[root].r || ed<node[root].l) return;
    if(st<=node[root].l && node[root].r<=ed) node[root].update(val);
    else
    {
        pushdown(root);
        update(root*2,st,ed,val);
        update(root*2+1,st,ed,val);
        pushup(root);
    }
}
LL query(int root,int st,int ed)
{
    if(ed<node[root].l || node[root].r<st) return INF;
    if(st<=node[root].l && node[root].r<=ed) return node[root].val;
    else
    {
        pushdown(root);
        LL lson=query(root*2,st,ed);
        LL rson=query(root*2+1,st,ed);
        pushup(root);
        return min(lson,rson);
    }
}


int main()
{
    cin>>n>>q;

    adjListInit(1,n);
    for(int i=1;i<=n-1;i++)
    {
        int a,b; LL c;
        scanf("%d%d%I64d",&a,&b,&c);
        addEdge(a,b,c,i);
    }

    dfsInit();
    dfs(1,-1);
    //for(int i=1;i<=n;i++) printf("in[%d]=%d  out[%d]=%d\n",i,in[i],i,out[i]);

    Eback.clear(); Eback_size=0;
    for(int i=n,a,b,c;i<=2*n-2;i++)
    {
        scanf("%d%d%d",&a,&b,&c);
        Eback.push_back(Edge(a,b,c)); Eback_size++;
        ptr[i]=Eback_size-1;
        Gback[a]=Eback_size-1;
    }
    Eback.push_back(Edge(1,1,0)); Eback_size++;
    Gback[1]=Eback_size-1;

    build(1,1,n);
    for(int i=1,type;i<=q;i++)
    {
        scanf("%d",&type);
        if(type==1)
        {
            int id; LL w;
            scanf("%d%I64d",&id,&w);
            if(id<n)
            {
                Edge &e=E[ptr[id]];
                //printf("Edge %d: %d->%d = %I64d\n",id,e.u,e.v,e.w);
                update(1,in[e.v],out[e.v],w-e.w);
                e.w=w;
            }
            else
            {
                Edge &e=Eback[ptr[id]];
                //printf("Edge %d: %d->%d = %I64d\n",id,e.u,e.v,e.w);
                update(1,in[e.u],in[e.u],w-e.w);
                e.w=w;
            }
        }
        if(type==2)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            if(in[u]<=in[v] && in[v]<=out[u]) //v在u统领的子树内
            {
                //printf("%d是%d的祖先\n",u,v);
                LL du=query(1,in[u],in[u])-Eback[Gback[u]].w;
                LL dv=query(1,in[v],in[v])-Eback[Gback[v]].w;
                printf("%I64d\n",dv-du);
            }
            else
            {
                //printf("%d不是%d的祖先\n",u,v);
                LL ans=query(1,in[u],out[u]);

                LL du=query(1,in[u],in[u])-Eback[Gback[u]].w;
                LL dv=query(1,in[v],in[v])-Eback[Gback[v]].w;
                ans-=du;
                ans+=dv;

                printf("%I64d\n",ans);
            }
        }
    }
}

PS.第一次200+行的代码1A，还是值得小小窃喜一下的

来源：oschina

链接：https://my.oschina.net/u/4320349/blog/3981941

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