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 25using 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;}