Traffic Network in Numazu HDU - 6393

http://acm.hdu.edu.cn/status.php?user=sunyutian&pid=6393

原图是一个ｎ条边的连通图　即一个树图之上又加了一条边　查询两点间的距离

并查集判出多余的边　用剩下的ｎ－１条边建树　对每次查询看两点之间的树上距离　以及经过多余边的距离　查距离用线段树或者树链剖分都可以

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define ll long long

struct node1
{
    int u;
    int v;
    ll w;
};

struct node2
{
    int v;
    int next;
};

struct node3
{
    int l;
    int r;
    ll val;
};

node1 pre[100010];
node2 edge[200010];
node3 tree[400010];
ll val[100010];
int f[100010],first[100010];
int deep[100010],fa[100010],sum[100010],son[100010],top[100010],mp1[100010],mp2[100010];
int n,q,num;

void addedge(int u,int v)
{
    edge[num].v=v;
    edge[num].next=first[u];
    first[u]=num++;
}//sunyutian

int getf(int p)
{
    if(f[p]==p) return p;
    else
    {
        f[p]=getf(f[p]);
        return f[p];
    }
}

bool unite(int u,int v)
{
    int fu,fv;
    fu=getf(u);
    fv=getf(v);
    if(fu!=fv)
    {
        f[fv]=fu;
        return true;
    }
    else return false;
}

void dfsI(int cur)
{
    int i,v;
    sum[cur]=1,son[cur]=-1;
    for(i=first[cur];i!=-1;i=edge[i].next)
    {
        v=edge[i].v;
        if(v!=fa[cur])
        {
            fa[v]=cur,deep[v]=deep[cur]+1;
            dfsI(v);
            sum[cur]+=sum[v];
            if(son[cur]==-1||sum[son[cur]]<sum[v])
            {
                son[cur]=v;
            }
        }
    }
}

void dfsII(int cur,int tp)
{
    int i,v;
    num++;
    top[cur]=tp,mp1[cur]=num,mp2[num]=cur;
    if(son[cur]==-1) return;
    dfsII(son[cur],tp);
    for(i=first[cur];i!=-1;i=edge[i].next)
    {
        v=edge[i].v;
        if(v!=fa[cur]&&v!=son[cur])
        {
            dfsII(v,v);
        }
    }
}

void pushup(int cur)
{
    tree[cur].val=tree[2*cur].val+tree[2*cur+1].val;
}

void build(int l,int r,int cur)
{
    int m;
    tree[cur].l=l;
    tree[cur].r=r;
    tree[cur].val=0;
    if(l==r)
    {
        tree[cur].val=val[mp2[l]];
        return;
    }
    m=(l+r)/2;
    build(l,m,2*cur);
    build(m+1,r,2*cur+1);
    pushup(cur);
}

void update(int tar,ll val,int cur)
{
    if(tree[cur].l==tree[cur].r)
    {
        tree[cur].val=val;
        return;
    }
    if(tar<=tree[2*cur].r) update(tar,val,2*cur);
    else update(tar,val,2*cur+1);
    pushup(cur);
}

ll query(int pl,int pr,int cur)
{
    ll res;
    if(pl<=tree[cur].l&&tree[cur].r<=pr) return tree[cur].val;
    res=0;
    if(pl<=tree[2*cur].r) res+=query(pl,pr,2*cur);
    if(pr>=tree[2*cur+1].l) res+=query(pl,pr,2*cur+1);
    return res;
}

ll solve(int u,int v)
{
    ll res;
    res=0;
    while(top[u]!=top[v])
    {
        if(deep[top[u]]<deep[top[v]]) swap(u,v);
        res+=query(mp1[top[u]],mp1[u],1);
        u=fa[top[u]];
    }
    if(u==v) return res;
    if(deep[u]<deep[v]) swap(u,v);
    res+=query(mp1[son[v]],mp1[u],1);
    return res;
}

int main()
{
    ll w,res1,res2,res3;
    int t,i,fu,fv,op,u,v,p;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&q);
        for(i=1;i<=n;i++) f[i]=i;
        memset(first,-1,sizeof(first));
        num=0,p=0;
        for(i=1;i<=n;i++)
        {
            scanf("%d%d%lld",&pre[i].u,&pre[i].v,&pre[i].w);
            if(unite(pre[i].u,pre[i].v))
            {
                addedge(pre[i].u,pre[i].v);
                addedge(pre[i].v,pre[i].u);
            }
            else p=i;
        }
        deep[1]=1,fa[1]=-1;
        dfsI(1);
        num=0;
        dfsII(1,1);
        for(i=1;i<=n;i++)
        {
            if(i!=p)
            {
                if(deep[pre[i].u]<deep[pre[i].v]) swap(pre[i].u,pre[i].v);
                val[pre[i].u]=pre[i].w;
            }
        }
        build(1,n,1);
        while(q--)
        {
            scanf("%d",&op);
            if(op==0)
            {
                scanf("%d%lld",&i,&w);
                if(i!=p) update(mp1[pre[i].u],w,1);
                else pre[i].w=w;
            }
            else
            {
                scanf("%d%d",&u,&v);
                res1=solve(u,v);
                res2=solve(u,pre[p].u)+solve(v,pre[p].v)+pre[p].w;
                res3=solve(u,pre[p].v)+solve(v,pre[p].u)+pre[p].w;
                printf("%lld\n",min(res1,min(res2,res3)));
            }
        }
    }
    return 0;
}