HDU 2586 how far away? (Beginner's LCA)

How far away? Time limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 7739 Accepted Submission (s): 2769


Problem Descriptionthere is n houses in the village and some bidirectional roads connecting them. Every day peole "What is it if I am want to go from House A to house B"? Usually it hard to answer. But luckily int this village the answer was always unique, since the roads was built in the the-the-the-that-there is a unique simp Le path ("simple" means you can ' t visit a place twice) between every the houses. Yout task is to answer all these curious people.
Inputfirst line was a single integer T (t<=10), indicating the number of test cases.
For each test case,in the first line there is the numbers N (2<=n<=40000) and M (1<=m<=200), the number of hous Es and the number of queries. The following n-1 lines each consisting three numbers i,j,k, separated bu a single space, meaning this there is a road con Necting House I and House j,with length K (0<k<=40000). The houses is labeled from 1 to N.
Next m lines each have distinct integers i and j, you areato answer the distance between House I and House J.
Outputfor each test case,output m lines. Each line represents the answer of the query. Output a bland line after all test case.
Sample Input

23 21 2 103 1 151 22 32 21 2 1001 22 1

Sample Output

1025100100

Approximate test instructions: Finding the distance between any two points in the tree is obviously the distance from the u,v to the root, minus the distance from their LCA to the root.

As a result of the study of LCA, three LCA methods have been used to find

Method one: Off-line algorithm Tarjan

#include <cstdio> #include <iostream> #include <cstring> #include <algorithm>using namespace std;const int N = 40000+1000;int head[n];int fst[n];struct edge{int v,w,nxt,id;} Es[n<<1],qury[444];int cnt,qc;int ans[222];int fa[n],ance[n];void inline add_query (int u,int v,int ID) {QURY[QC].    V=v;    Qury[qc].id=id;    Qury[qc].nxt=fst[u];    fst[u]=qc++;    Qury[qc].v=u;    Qury[qc].id=id;    QURY[QC].NXT=FST[V]; fst[v]=qc++;}    void inline Add_edge (int u,int v,int W) {es[cnt].w=w;    Es[cnt].v=v;    Es[cnt].nxt=head[u];    head[u]=cnt++;    Es[cnt].v=u;    Es[cnt].w=w;    ES[CNT].NXT=HEAD[V]; head[v]=cnt++;} int N,m;bool vis[n];int GETF (int x) {return x==fa[x]? X:FA[X]=GETF (Fa[x]);} void Merge (int u,int v) {fa[getf (U)]=getf (v);}    void LCA (int u,int pa) {ance[u]=fa[u]=u;        for (int i=head[u];~i;i=es[i].nxt) {int v=es[i].v;        if (V==PA) continue;        LCA (V,u);        Merge (U,V);    ANCE[GETF (v)]=u;    } vis[u]=1; for (int i=fST[U];~I;I=QURY[I].NXT) {int v=qury[i].v;    if (Vis[v]) ans[qury[i].id]=ance[getf (v)];        }}int dp[n];void MAKEDP (int u,int pa) {for (int i=head[u];~i;i=es[i].nxt) {int v=es[i].v,w=es[i].w;        if (V==PA) continue;        Dp[v]=dp[u]+w;    MAKEDP (V,u);    }}void Ini () {memset (head,-1,sizeof (head));    memset (fst,-1,sizeof (FST));    cnt=qc=0; memset (vis,0,sizeof (Vis));}    int main () {int T;    scanf ("%d", &t);        while (t--) {ini ();        scanf ("%d%d", &n,&m);            for (int i=1;i<n;i++) {int u,v,w;            scanf ("%d%d%d", &u,&v,&w);        Add_edge (U,V,W);            } for (int i=1;i<=m;i++) {int u,v;            scanf ("%d%d", &u,&v);        Add_query (U,v,i);        } LCA (a);        MAKEDP (a);            for (int i=0;i<m;i++) {int j=i<<1;        printf ("%d\n", Dp[qury[j].v]+dp[qury[j^1].v]-2*dp[ans[i+1]]);  }  } return 0;} 

Method Two: BFs Multiplication hop Table algorithm

#include <cstdio> #include <iostream> #include <cstring> #include <queue> #include < algorithm>using namespace Std;const int N = 40000+1000;int head[n];struct edge{int v,w,nxt;}    Es[n<<1];int cnt;inline void Add_edge (int u,int v,int W) {es[cnt].v=v;    Es[cnt].w=w;    Es[cnt].nxt=head[u];    head[u]=cnt++;    Es[cnt].v=u;    Es[cnt].w=w;    ES[CNT].NXT=HEAD[V]; head[v]=cnt++;}    int dp[n];int dep[n];bool vis[n];int pa[n][20];void bfs () {queue<int>q;    Q.push (1);    Pa[1][0]=1;    Vis[1]=1;        while (!q.empty ()) {int U=q.front (); Q.pop ();        for (int i=1;i<20;i++) pa[u][i]=pa[pa[u][i-1]][i-1];            for (int i=head[u];~i;i=es[i].nxt) {int v=es[i].v,w=es[i].w;                if (vis[v]==0) {vis[v]=1;                Dp[v]=dp[u]+w;                Pa[v][0]=u;                dep[v]=dep[u]+1;            Q.push (v); }}}}int LCA (int u,int v) {if (Dep[u]>dep[v]) Swap (U, v);    for (int det=dep[v]-dep[u],i=0;det;i++,det>>=1) if (det&1) v=pa[v][i];    if (v==u) return v;    for (int i=20-1;i>=0;i--) if (Pa[u][i]!=pa[v][i]) v=pa[v][i],u=pa[u][i]; return pa[u][0];}    int n,m;void ini () {memset (head,-1,sizeof (head));    cnt=0;    memset (vis,0,sizeof (VIS)); dp[1]=0;}    int main () {int T;    scanf ("%d", &t);        while (t--) {scanf ("%d%d", &n,&m);        INI ();            for (int i=1;i<n;i++) {int u,v,w;            scanf ("%d%d%d", &u,&v,&w);        Add_edge (U,V,W);        } BFS ();            for (int i=1;i<=m;i++) {int u,v;            scanf ("%d%d", &u,&v);            int LCAV = LCA (u,v);            int ans = DP[U]+DP[V]-2*DP[LCAV];        printf ("%d\n", ans); }} return 0;}

Method Three: DFS-based RMQ

#include <cstdio> #include <iostream> #include <cstring> #include <queue> #include < algorithm>using namespace Std;const int N = 40000+1000;int head[n];struct edge{int v,w,nxt;}    Es[n<<1];int cnt;inline void Add_edge (int u,int v,int W) {es[cnt].v=v;    Es[cnt].w=w;    Es[cnt].nxt=head[u];    head[u]=cnt++;    Es[cnt].v=u;    Es[cnt].w=w;    ES[CNT].NXT=HEAD[V]; head[v]=cnt++;}    int dp[n];int index;int vs[n*3],id[n],dep[n];int lca[n*3][20];bool vis[n];void dfs (int u,int h) {vis[u]=1;    Id[u]=++index;    Vs[index]=u;    Dep[u]=h;        for (int i=head[u];~i;i=es[i].nxt) {int v=es[i].v,w=es[i].w;        if (Vis[v]) continue;        Dp[v]=dp[u]+w;        DFS (V,H+1);    Vs[++index]=u;    }}int mm[n];void Ini () {memset (vis,0,sizeof (VIS));    memset (head,-1,sizeof (head));    cnt=index=0; dp[1]=0;}    int N,m;int Main () {mm[0]=-1; for (int i=1;i<=n-1;i++) mm[i]= (((i-1) &i) ==0)?    MM[I-1]+1:MM[I-1];    int T;    scanf ("%d", &t); WhiLe (t--) {scanf ("%d%d", &n,&m);        INI ();            for (int i=1;i<n;i++) {int u,v,w;            scanf ("%d%d%d", &u,&v,&w);        Add_edge (U,V,W);        } dfs (1,0);        for (int i=1;i<=index;i++) lca[i][0]=vs[i]; for (int j=1;j<20;j++) for (int i=1;i+ (1&LT;&LT;J) -1<=index;i++) {int a=lca[i][j                -1],b=lca[i+ (1<< (j-1))][j-1]; LCA[I][J] = dep[a]<dep[b]?            A:B;            } for (int i=1;i<=m;i++) {int u,v;            scanf ("%d%d", &u,&v);            int l=min (id[u],id[v]);            int R=max (id[u],id[v]);            int k=mm[r-l+1];            int a=lca[l][k],b=lca[r-(1<<k) +1][k]; int lcav = Dep[a]<dep[b]?            A:B;            int ans = DP[U]+DP[V]-2*DP[LCAV];        printf ("%d\n", ans); }} return 0;}

Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced.

HDU 2586 how far away? (Beginner's LCA)