BZOJ 2125: 最短路 （仙人掌，树链剖分）

第一道仙人掌题.   

由于仙人掌中每条边最多只属于一个环，所以两个在环中的点的最短距离是好算的.    

code: 

#include <bits/stdc++.h> 
#define N 200006 
#define setIO(s) freopen(s".in","r",stdin) 
using namespace std;  
int edges,n,m,Q,tim,tot;     
struct Edge 
{
    int v,c;    
    Edge(int v=0,int c=0):v(v),c(c){}  
};    
vector<Edge>G[N];    
int hd[N],to[N<<1],nex[N<<1],fa[N],low[N],dfn[N],val[N],dis[N],cdis[N];            
int dep[N],size[N],top[N],son[N],point[N];      
void add(int u,int v,int c) 
{
    nex[++edges]=hd[u],hd[u]=edges,to[edges]=v,val[edges]=c;   
}            
void build(int ff,int x,int c) 
{
    for(int i=x;i!=ff;i=fa[i])         
        cdis[i]=c,c+=dis[i];
    cdis[++tot]=c;               
    G[tot].push_back(Edge(ff,0));       
    G[ff].push_back(Edge(tot,0));    
    for(int i=x;i!=ff;i=fa[i])  
    {
        G[tot].push_back(Edge(i,min(cdis[i],c-cdis[i])));      
        G[i].push_back(Edge(tot,min(cdis[i],c-cdis[i])));    
    }   
}
void tarjan(int x,int ff) 
{
    fa[x]=ff;     
    low[x]=dfn[x]=++tim;      
    for(int i=hd[x];i;i=nex[i]) 
    {
        int y=to[i];  
        if(y==ff) continue;     
        if(!dfn[y]) 
            dis[y]=val[i],tarjan(y,x),low[x]=min(low[x],low[y]);
        else low[x]=min(low[x],dfn[y]);    
        if(low[y]>dfn[x])  
        {
            G[x].push_back(Edge(y,val[i]));
            G[y].push_back(Edge(x,val[i]));  
        }
    }
    for(int i=hd[x];i;i=nex[i]) 
    {
        int y=to[i];    
        if(fa[y]!=x&&dfn[y]>dfn[x])    
            build(x,y,val[i]);   
    }
}
void dfs1(int x,int ff) 
{     
    fa[x]=ff;    
    for(int i=0;i<G[x].size();++i) 
    {
        int y=G[x][i].v,c=G[x][i].c; 
        if(y==ff)  
            continue;    
        dis[y]=dis[x]+c;   
        dep[y]=dep[x]+1;    
        dfs1(y,x);   
        size[x]+=size[y];   
        if(size[y]>size[son[x]])   
            son[x]=y;     
    }           
}
void dfs2(int x,int tp) 
{
    top[x]=tp;  
    point[dfn[x]=++tim]=x;       
    if(son[x])  
        dfs2(son[x],tp);  
    for(int i=0;i<G[x].size();++i) 
        if(G[x][i].v!=fa[x]&&G[x][i].v!=son[x])   
            dfs2(G[x][i].v,G[x][i].v);   
}
int get_lca(int x,int y) 
{
    while(top[x]!=top[y])   
        dep[top[x]]>dep[top[y]]?x=fa[top[x]]:y=fa[top[y]];  
    return dep[x]<dep[y]?x:y;    
}
int jump(int x,int lca) 
{
    int las;   
    while(top[x]!=top[lca]) 
        las=top[x],x=fa[top[x]];    
    return x==lca?las:point[dfn[lca]+1];   
}
int main() 
{ 
    // setIO("input");      
    scanf("%d%d%d",&n,&m,&Q);    
    for(int i=1;i<=m;++i) 
    {
        int u,v,c;   
        scanf("%d%d%d",&u,&v,&c);   
        add(u,v,c),add(v,u,c);   
    }       
    tot=n,tarjan(1,0);             
    dis[1]=dep[1]=tim=0;      
    dfs1(1,0);       
    dfs2(1,1);  
    for(int i=1;i<=Q;++i) 
    {
        int x,y; 
        scanf("%d%d",&x,&y);   
        int lca=get_lca(x,y);   
        if(lca<=n) 
            printf("%d\n",dis[x]+dis[y]-(dis[lca]<<1));  
        else 
        {
            int p1=jump(x,lca);   
            int p2=jump(y,lca);    
            int ans=dis[x]-dis[p1]+dis[y]-dis[p2]+min(abs(cdis[p2]-cdis[p1]),cdis[lca]-abs(cdis[p2]-cdis[p1]));  
            printf("%d\n",ans);   
        }
    }
    return 0;  
}