poj 1986 tarjan/rmq(LCA问题)

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 40005
struct 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 40005
struct 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;
}