算法模板之最近公共祖先问题（LCA）

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poj1330

最近公共祖先问题模板：

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<cstdlib>#include<string>#include<vector>#define N 30000using namespace std;int id[N],vis[N],depth[N],in[N],dp[N][20],k;vector<int>G[N];int Min(int i,int j){    if(depth[i]<=depth[j])        return i;    return j;}void rmq_init(int n){    for(int i=1; i<=n; i++)        dp[i][0]=i;    for(int j=1; (1<<j)<=n; j++)        for(int i=1; i+(1<<j)-1<=n; i++)            dp[i][j]=Min(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);}void dfs(int v,int d){    id[v]=k;    vis[k]=v;    depth[k]=d;    k++;    for(int i=0; i<G[v].size(); i++)    {        dfs(G[v][i],d+1);        vis[k]=v;        depth[k]=d;        k++;    }}void init(int root,int n){    k=1;    dfs(root,0);    rmq_init(n*2-1);}int query(int l,int r){    int k=(int)(log10(r-l+1)/log10(2.0));    return Min(dp[l][k],dp[r-(1<<k)+1][k]);}int main(){    int t,n,a,b,x,y;    cin>>t;    while(t--)    {        for(int i=0; i<N; i++)            G[i].clear();        memset(in,0,sizeof(in));        cin>>n;        for(int i=1; i<n; i++)        {            scanf("%d%d",&a,&b);            G[a].push_back(b);            in[b]=1;        }        int i;        for(i=1; i<=n; i++)            if(in[i]==0)                break;        init(i,n);        cin>>x>>y;        int ans=query(min(id[x],id[y]),max(id[x],id[y]));        printf("%d\n",vis[ans]);    }    return 0;}

poj1986

求解树上两点相距的最近距离模板：

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<cstdlib>#include<string>#include<vector>#define N 40010using namespace std;int vis[2*N],depth[2*N],id[N],dis[N],dp[N*2][20],k;struct node{    int to,w;};vector<node>G[N];int Min(int a,int b){    if(depth[a]<=depth[b])        return a;    return b;}void rmq_init(int n){    for(int i=1; i<=n; i++)        dp[i][0]=i;    for(int j=1; (1<<j)<=n; j++)        for(int i=1; i+(1<<j)-1<=n; i++)            dp[i][j]=Min(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);}void dfs(int v,int p,int d){    id[v]=k;    vis[k]=v;    depth[k]=d;    k++;    for(int i=0; i<G[v].size(); i++)    {        if(G[v][i].to!=p)        {            dis[G[v][i].to]=dis[v]+G[v][i].w;            dfs(G[v][i].to,v,d+1);            vis[k]=v;            depth[k]=d;            k++;        }    }}void init(int n){    k=1;    dfs(1,-1,0);    rmq_init(2*n-1);}int query(int l,int r){    int k=(int)(log10(r-l+1)/log10(2.0));    int num=Min(dp[l][k],dp[r-(1<<k)+1][k]);    return vis[num];}int main(){    int n,m,a,b,c,q,x,y;    char ch[10];    while(cin>>n>>m)    {        for(int i=0; i<N; i++)        {            G[i].clear();            id[i]=0;            dis[i]=0;        }        for(int i=0; i<m; i++)        {            scanf("%d%d%d%s",&a,&b,&c,ch);            node nd;            nd.to=b,nd.w=c;            G[a].push_back(nd);            nd.to=a,nd.w=c;            G[b].push_back(nd);        }        init(n);        cin>>q;        while(q--)        {            scanf("%d%d",&x,&y);            int ans=query(min(id[x],id[y]),max(id[x],id[y]));            printf("%d\n",dis[x]+dis[y]-2*dis[ans]);        }    }    return 0;}

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