POJ.1986 Distance Queries ( LCA 倍增 )

题意分析

给出一个N个点，M条边的信息(u,v,w),表示树上u-v有一条边，边权为w，接下来有k个询问，每个询问为(a,b)，求a,b两点到lca(a,b)的边权之和为多少。

倍增维护树上前缀和，求得LCA之后，相应做差即可。

代码总览

#include <cstdio>

#include <algorithm>

#include <cstring>

#include <cmath>

#define nmax 80520

#define demen 25

using namespace std;

int fa[nmax][demen],dis[nmax],head[nmax],dep[nmax];

int n,m,tot = 0;

struct node{

    int to;

    int next;

    int w;

}edge[nmax];

void add(int u, int v, int w){

    edge[tot].to = v;

    edge[tot].next = head[u];

    edge[tot].w = w;

    head[u] = tot++;

}

void dfs(int rt,int f){

    fa[rt][0] = f;

    for(int i = 1;i<=20;++i){

        fa[rt][i] = fa[fa[rt][i-1]][i-1];

    }

    for(int i = head[rt];i!=-1;i = edge[i].next){

        int nxt = edge[i].to;

        if(nxt != f){

            dis[nxt] = dis[rt] + edge[i].w;

            dep[nxt] = dep[rt] + 1;

            dfs(nxt,rt);

        }

    }

}

int lca(int x, int y){

    int X = x,Y=y;

    if(dep[x] < dep[y]) swap(x,y);

    for(int i = 20;i>=0;--i){

        if(dep[y] <= dep[fa[x][i]])

            x = fa[x][i];

    }

    if(x == y) return(abs(dis[X] - dis[Y]));

    for(int i = 20;i>=0;--i){

        if(fa[x][i] != fa[y][i]){

            x = fa[x][i],y = fa[y][i];

        }

    }

    return(dis[X]+dis[Y] - 2*dis[fa[x][0]]);

}

void init(){

    memset(fa,0,sizeof fa);

    memset(head,-1,sizeof head);

    memset(dis, 0, sizeof dis);

    memset(dep,0,sizeof dep);

    tot = 0;

}

int main()

{

    while(scanf("%d %d",&n,&m) != EOF){

        init();

        int u,v,w,x,y;

        char c;

        for(int i = 0;i<m;++i){

            scanf("%d %d %d %c",&u,&v,&w,&c);

            add(u,v,w);

            add(v,u,w);

        }

        dep[1] = 1;

        dfs(1,0);

        int k = 0; scanf("%d",&k);

        for(int i = 0;i<k;++i){

            scanf("%d %d",&x,&y);

            printf("%d\n",lca(x,y));

        }

    }

    return 0;

}