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zcmimi
2019-01-30

建树

tarjan求出双连通分量后把其中的点连向新建的节点

建成的新图(树)即为圆方树

性质:可以发现原图中两点间必经的点就是圆方树中的圆点

道路相遇

可以用找出lca后直接求

#include<bits/stdc++.h>
#define il __inline__ __attribute__ ((always_inline))
#define MB template <class T>il
#define Fur(i,x,y) for(int i(x);i<=y;++i)
#define fl(i,x) for(int i(head[x]),to;to=e[i].to,i;i=e[i].nxt)
#define FL(i,x) for(int i(HEAD[x]),to;to=e[i].to,i;i=e[i].nxt)
MB T MAX(T x,T y){return x>y?x:y;}
MB T MIN(T x,T y){return x<y?x:y;}
using namespace std;
namespace IO{const int str=1<<20;static char in_buf[str],*in_s,*in_t;bool __=0;il char gc(){return (in_s==in_t)&&(in_t=(in_s=in_buf)+fread(in_buf,1,str,stdin)),in_s==in_t?__=1,EOF:*in_s++;}il void in(string &ch){ch.clear();if(__)return;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF){__=1;return;}ch+=c;while((c=gc())!=EOF&&!isspace(c))ch+=c;if(c==EOF)__=1;}il void in(char &ch){if(__)return;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)__=1;else ch=c;}il void in(char *ch){*ch='\0';if(__)return;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF){__=1;return;}*ch=c;ch++;while((c=gc())!=EOF&&!isspace(c))*ch=c,ch++;if(c==EOF)__=1;*ch='\0';}template<typename T>il void in(T &x){if(__)return;char c=gc();bool f=0;while(c!=EOF&&(c<'0'||c>'9'))f^=(c=='-'),c=gc();if(c==EOF){__=1;return;}x=0;while(c!=EOF&&'0'<=c&&c<='9')x=x*10+c-48,c=gc();if(c==EOF)__=1;if(f)x=-x;}template<typename T,typename ... arr>il void in(T &x,arr & ... y){in(x),in(y...);}const char ln='\n';static char out_buf[str],*out_s=out_buf,*out_t=out_buf+str;il void flush(){fwrite(out_buf,1,out_s-out_buf,stdout);out_s=out_buf;}il void pt(char c){(out_s==out_t)?(fwrite(out_s=out_buf,1,str,stdout),*out_s++=c):(*out_s++=c);}il void out(const char* s){while(*s)pt(*s++);}il void out(char* s){while(*s)pt(*s++);}il void out(char c){pt(c);}il void out(string s){for(int i=0;s[i];i++)pt(s[i]);}template<typename T>il void out(T x){if(!x){pt('0');return;}if(x<0)pt('-'),x=-x;char a[50],t=0;while(x)a[t++]=x%10,x/= 10;while(t--)pt(a[t]+'0');}template<typename T,typename ... arr>il void out(T x,arr & ... y){out(x),out(y...);}}using namespace IO;
const int N=1000011;
int n,m,cnt=0,head[N],HEAD[N];
struct edge{int to,nxt;}e[N<<3];
il void add(int x,int y){
    e[++cnt].to=y;e[cnt].nxt=head[x];head[x]=cnt;
    e[++cnt].to=x;e[cnt].nxt=head[y];head[y]=cnt;
}
il void ADD(int x,int y){
    e[++cnt].to=y;e[cnt].nxt=HEAD[x];HEAD[x]=cnt;
    e[++cnt].to=x;e[cnt].nxt=HEAD[y];HEAD[y]=cnt;
}
int dfn[N],low[N],sz=0,tot,st[N],tp=0;
void tarjan(int x,int fa){
    dfn[x]=low[x]=++sz;st[++tp]=x;
    fl(i,x)
    if(!dfn[to]){
        tarjan(to,x);
        low[x]=MIN(low[x],low[to]);
        if(low[to]>=dfn[x]){
            ADD(++tot,x);
            while(int k=st[tp--]){
                ADD(tot,k);
                if(k==to)break;
            }
        }
    }
    else if(to!=fa)low[x]=MIN(low[x],dfn[to]);
}
int f[N],siz[N],d[N],top[N],son[N];
void dfs(int x){
    siz[x]=1;
    FL(i,x)if(to!=f[x]){
        f[to]=x;
        d[to]=d[x]+1;
        dfs(to);
        siz[x]+=siz[to];
        if(siz[to]>siz[son[x]])son[x]=to;
    }
}
void bt(int x,int tp){
    top[x]=tp;
    if(son[x])bt(son[x],tp);
    FL(i,x)if(!top[to])bt(to,to);
}
int lca(int x,int y){
    while(top[x]!=top[y])
        (d[top[x]]>d[top[y]])?x=f[top[x]]:y=f[top[y]];
    return d[x]<d[y]?x:y;
}
int main(){
    in(n,m);tot=n;
    int x,y,q;
    Fur(i,1,m)in(x,y),add(x,y);
    tarjan(1,0);
    d[1]=1;dfs(1);bt(1,1);
    in(q);
    while(q--){
        in(x,y);
        out((d[x]+d[y]-2*d[lca(x,y)])/2+1,ln);
    }
    flush();
}

圆方树与仙人掌

BZOJ 2125 最短路

LG 5236 静态仙人掌

还是先建成圆方树

若两个点的lca是原图的点,那么直接d_x+d_y-2\times d_{lca(x,y)}

否则就是两个点到环(方点)的距离加上两个点在环上的最短距离(可以用前缀和解决)

#include<cstdio>
#pragma GCC optimize(3)
#define il __inline__ __attribute__ ((always_inline))
#define Fur(i,x,y) for(int i(x);i<=y;++i)
#define Fdr(i,x,y) for(int i(x);i>=y;--i)
il void swap(int &x,int &y){int t=x;x=y;y=t;}
il int min(int x,int y){return x<y?x:y;}
il int abs(int x){return x>0?x:-x;}
namespace IO{const int str=1<<20;static char in_buf[str],*in_s,*in_t;bool __=0;il char gc(){return (in_s==in_t)&&(in_t=(in_s=in_buf)+fread(in_buf,1,str,stdin)),in_s==in_t?__=1,EOF:*in_s++;}template<typename T>il void in(T &x){if(__)return;char c=gc();bool f=0;while(c!=EOF&&(c<'0'||c>'9'))f^=(c=='-'),c=gc();if(c==EOF){__=1;return;}x=0;while(c!=EOF&&'0'<=c&&c<='9')x=x*10+c-48,c=gc();if(c==EOF)__=1;if(f)x=-x;}template<typename T,typename ... arr>il void in(T &x,arr & ... y){in(x),in(y...);}const char ln='\n';static char out_buf[str],*out_s=out_buf,*out_t=out_buf+str;il void flush(){fwrite(out_buf,1,out_s-out_buf,stdout);out_s=out_buf;}il void pt(char c){(out_s==out_t)?(fwrite(out_s=out_buf,1,str,stdout),*out_s++=c):(*out_s++=c);}il void out(const char* s){while(*s)pt(*s++);}il void out(char* s){while(*s)pt(*s++);}il void out(char c){pt(c);}template<typename T>il void out(T x){if(!x){pt('0');return;}if(x<0)pt('-'),x=-x;char a[50],t=0;while(x)a[t++]=x%10,x/= 10;while(t--)pt(a[t]+'0');}template<typename T,typename ... arr>il void out(T x,arr & ... y){out(x),out(y...);}}using namespace IO;
const int N=20011;
int n,m,q,cnt=0,head[N],HEAD[N];
struct edge{
    int to,nxt,w;
}e[N<<2];
#define fl(i,x) for(int i(head[x]),to;to=e[i].to,i;i=e[i].nxt)
#define FL(i,x) for(int i(HEAD[x]),to;to=e[i].to,i;i=e[i].nxt)
il void add(int x,int y,int w){
    e[++cnt].to=y;e[cnt].nxt=head[x];head[x]=cnt;e[cnt].w=w;
    e[++cnt].to=x;e[cnt].nxt=head[y];head[y]=cnt;e[cnt].w=w;
}
il void ADD(int x,int y,int w){
    e[++cnt].to=y;e[cnt].nxt=HEAD[x];HEAD[x]=cnt;e[cnt].w=w;
    e[++cnt].to=x;e[cnt].nxt=HEAD[y];HEAD[y]=cnt;e[cnt].w=w;
}
int dfn[N],low[N],sz=0,fa[N],tot,b[N],id[N];
int s[N],dep[N];
void solve(int x,int y,int w){
    ++tot;
    int t=w,tt=0;
    for(int i=y;i!=fa[x];i=fa[i]){
        s[i]=t;
        t+=b[i];
        id[i]=tt++;
    }
    s[tot]=s[x];s[x]=0;
    for(int i=y;i!=fa[x];i=fa[i])
        ADD(tot,i,min(s[i],s[tot]-s[i]));
}
void tarjan(int x){
    dfn[x]=low[x]=++sz;
    fl(i,x)if(to!=fa[x]){
        if(!dfn[to]){
            fa[to]=x;b[to]=e[i].w;
            tarjan(to);
            low[x]=min(low[x],low[to]);
        }
        else low[x]=min(low[x],dfn[to]);
        if(low[to]>dfn[x])ADD(x,to,e[i].w);
    }
    fl(i,x)if(fa[to]!=x&&dfn[to]>dfn[x])
        solve(x,to,e[i].w);
}
int f[16][N],d[N],A,B;
void dfs(int x){
    for(int j=1;(1<<j)<=d[x];++j)
        f[j][x]=f[j-1][f[j-1][x]];
    FL(i,x)if(to!=f[0][x]){
        f[0][to]=x;
        d[to]=d[x]+1;
        dep[to]=dep[x]+e[i].w;
        dfs(to);
    }
}
int lca(int x,int y){
    if(d[x]<d[y])swap(x,y);
    int D=d[x]-d[y];
    for(int i=0;(1<<i)<=D;++i)if(D&(1<<i))x=f[i][x];
    if(x==y)return x;
    Fdr(i,15,0)if((1<<i)<=d[x]&&f[i][x]!=f[i][y])
        x=f[i][x],y=f[i][y];
    A=x,B=y;
    return f[0][x];
}
int main(){
    in(n,m,q);tot=n;
    int x,y,w;
    Fur(i,1,m)in(x,y,w),add(x,y,w);
    tarjan(1);
    dfs(1);
    while(q--){
        in(x,y);
        w=lca(x,y);
        if(w<=n)out(dep[x]+dep[y]-2*dep[w],ln);
        else out(dep[x]+dep[y]-dep[A]-dep[B]+min(s[w]-abs(s[B]-s[A]),abs(s[A]-s[B])),ln);
    }
    flush();
}

仙人掌直径

圆方树
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