poj 1986 tarjan/rmq(LCA问题)

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题意：给出一棵树，求任意两点的最小距离。查询为多组。

思路：LCA(Least Common Ancestor最近公共祖先)。设LCA(X, Y) = L，dist(X)表示X到根节点的距离（Y同理），那么X到Y的路径长度就是dist(X) + dist(Y) - 2 * dist(L)。

离线方法为tarjan算法，本质是深搜+并查集。复杂度O(m+q)，q为查询的对数。

在线方法为rmq，（具体算法参见http://ayzk.wordpress.com.cn/archives/14），复杂度为O(nlogn)+O(q).

tarjan算法：

#include <stdio.h>#include <string.h>#define N 10005#define M 40005struct edge{int y,w,next;}e[M<<1];struct query{int y,index,next;}q[N<<1];int res[N],firstq[M],firste[M],visited[M],root[M],dis2root[M];int n,m,top_q,top_e;void init(){int i;top_q = top_e = 0;memset(res,0,sizeof(res));memset(firstq,-1,sizeof(firstq));memset(firste,-1,sizeof(firste));memset(visited,0,sizeof(visited));memset(dis2root,0,sizeof(dis2root));for(i = 0;i<=n;i++)root[i] = i;}void adde(int x,int y,int w){e[top_e].y = y;e[top_e].w = w;e[top_e].next = firste[x];firste[x] = top_e++;}void addq(int x,int y,int index){q[top_q].y = y;q[top_q].index = index;q[top_q].next = firstq[x];firstq[x] = top_q++;}int find(int x){if(root[x] == x)return x;else return root[x] = find(root[x]);}void dfs(int x,int dis){int j,y;visited[x] = 1;for(j = firste[x];j!=-1;j=e[j].next){y = e[j].y;if(!visited[y]){dis2root[y] = dis+e[j].w;dfs(y,dis+e[j].w);root[y] = x;}}for(j = firstq[x];j!=-1;j=q[j].next)if(visited[q[j].y]){res[q[j].index] = dis2root[x]+dis2root[q[j].y]-2*dis2root[find(q[j].y)];}}int main(){freopen("a.txt","r",stdin);while(scanf("%d %d",&n,&m)!=EOF){int i,j,a,b,w;char ch;init();for(i = 0;i<m;i++){scanf("%d %d %d %c",&a,&b,&w,&ch);adde(a,b,w);adde(b,a,w);}scanf("%d",&m);for(i = 0;i<m;i++){scanf("%d %d",&a,&b);addq(a,b,i);addq(b,a,i);}dfs(1,0);for(i = 0;i<m;i++)printf("%d\n",res[i]);}return 0;}

rmq算法：

#include <stdio.h>#include <string.h>#include <math.h>#define min(a,b) ((a)<(b)?(a):(b))#define clc(s,t) memset(s,t,sizeof(s))#define swap(a,b,k) {k=a;a=b;b=k;}#define N 40005struct edge{    int y,w,next;}e[N*2];int first[N],top,n,m,q,dis[N];int flag[N<<1],r[N],d[N<<1],dp[N<<1][20],len;void init(){    clc(first,-1);    clc(r, 0);    clc(dis, 0);    top = len = 0;}void add(int x,int y,int w){    e[top].y = y;    e[top].w = w;    e[top].next = first[x];    first[x] = top++;}void dfs(int x,int fa,int dep){    int i;    for(i = first[x];i!=-1;i=e[i].next)        if(e[i].y != fa){            dis[e[i].y] = dis[x]+e[i].w;            flag[++len] = e[i].y;            r[e[i].y] = len;            d[len] = dep+1;            dfs(e[i].y,x,dep+1);            flag[++len] = x;            d[len] = dep;        }}void st(int n){    int i,j;    int k = log((double)(1+n))/log(2.0);    for(i = 1;i<=n;i++)        dp[i][0] = i;    for(j = 1;j<=k;j++)        for(i = 1;i+(1<<j)-1<=n;i++){            if(d[dp[i][j-1]] < d[dp[i+(1<<(j-1))][j-1]])                dp[i][j] = dp[i][j-1];            else                dp[i][j] = dp[i+(1<<(j-1))][j-1];        }}int query(int a,int b){    int k = log((double)(b-a+1))/log(2.0);    if(d[dp[a][k]] < d[dp[b-(1<<k)+1][k]])        return dp[a][k];    return dp[b-(1<<k)+1][k];}int main(){    int i,a,b,w,lca;    char ch;    init();    scanf("%d %d",&n,&m);    for(i = 0;i<m;i++){        scanf("%d %d %d %c",&a,&b,&w,&ch);        add(a,b,w);        add(b,a,w);    }    scanf("%d",&q);    r[1] = flag[++len] = 1;    dfs(1,-1,0);    st(n*2-1);    while(q--){        scanf("%d %d",&a,&b);        if(r[a]>r[b]){            swap(a,b,w);        }        lca = flag[query(r[a], r[b])];        printf("%d\n",dis[a]+dis[b]-2*dis[lca]);            }    return 0;}

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