POJ.1986 Distance Queries ( LCA 倍增 )

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;
}