JZOI 距离 (Standard IO) 题解

洛谷AC通道

题目描述

给出一棵树，求树上两点间的距离

考虑用树链剖分+线段树做。将边权下移，变成点的权值，然后统计点权和即可。

当然，注意，在统计点权时，实际上LCA这个点的权值并不包含在我们的距离之内，因此需要减掉。

#include <bits/stdc++.h>
using namespace std;
#define N 10010

inline int read(){
    int x = 0, s = 1;
    char c = getchar();
    while(!isdigit(c)){
        if(c == '-') s = -1;
        c = getchar();
    }
    while(isdigit(c)){
        x = x * 10 + (c ^ '0');
        c = getchar(); 
    }
    return x * s;
}

struct node{
    int v, w;
    int next;
} t[N << 1];
int f[N];

int bian = 0;
inline void add(int u, int v, int w){
    t[++bian] = (node){v, w, f[u]}, f[u] = bian;
    t[++bian] = (node){u, w, f[v]}, f[v] = bian;
    return ;
}  

int dfn[N], cnt = 0;
int fa[N], son[N], top[N], siz[N], deth[N];
int a[N];
int n, m;

/*---------树链剖分start--------*/
#define v t[i].v
void dfs1(int now, int father){
    deth[now] = deth[father] + 1;
    siz[now] = 1;
    fa[now] = father;
    for(int i = f[now]; i; i = t[i].next){
        if(v != father){
            dfs1(v, now);
            siz[now] += siz[v];
            if(siz[v] > siz[son[now]])
                son[now] = v;
        }
    }
    return ;
}

void dfs2(int now, int tp){
    top[now] = tp;
    dfn[now] = ++cnt;
    if(!son[now]) return ;
    dfs2(son[now], tp);
    for(int i = f[now]; i; i = t[i].next){
        if(v != fa[now] && v != son[now])
            dfs2(v, v);
    }
    return ;
}

void dfs3(int now, int father){
    for(int i = f[now]; i; i = t[i].next){
        if(v != father){
            a[dfn[v]] = t[i].w;
            dfs3(v, now);
        }
    }
    return ;
}
#undef v

int lca(int x, int y){
    while(top[x] != top[y]){
        if(deth[top[x]] < deth[top[y]]) swap(x, y);
        x = fa[top[x]]; 
    } 
    return deth[x] < deth[y] ? x : y;
}
/*-----------end-------*/

/*---------线段树start--------*/

struct tree{
    int w;
} e[N << 2];

inline void pushup(int o){
    e[o].w = e[o << 1].w + e[o << 1 | 1].w;
    return ;
}

void build(int o, int l, int r){
    if(l == r){
        e[o].w = a[l];
        return ;
    }
    int mid = l + r >> 1;
    build(o << 1, l, mid);
    build(o << 1 | 1, mid + 1, r);
    pushup(o);
    return ;
}

int query(int o, int l, int r, int in, int end){
    if(l > end || r < in) return 0;
    if(l >= in && r <= end){
        return e[o].w;
    }
    int mid = l + r >> 1;
    return query(o << 1, l, mid, in, end) + query(o << 1 | 1, mid + 1, r, in, end);
}

int ask_he(int x, int y){
    int sum = 0;
    while(top[x] != top[y]){
        if(deth[top[x]] < deth[top[y]]) swap(x, y);
        sum += query(1, 1, n, dfn[top[x]], dfn[x]);
        x = fa[top[x]];
    }
    if(deth[x] > deth[y]) swap(x, y);
    sum += query(1, 1, n, dfn[x], dfn[y]);
    return sum;
}
/*------------线段树end------------*/

inline void clean(){
    memset(son, 0, sizeof(son));
    memset(dfn, 0, sizeof(dfn));
    cnt = 0;
    for(int i = 1;i <= bian; i++)
        t[i] = (node){0, 0, 0};
    memset(f, 0, sizeof(f));
    bian = 0;
    return ; 
}

int main(){
//    freopen("10.in", "r", stdin);
    int T = read();
    while(T--){
        clean();
        n = read(), m = read();
        for(int i = 1;i < n; i++){
            int x = read(), y = read(), w = read();
            add(x, y, w);
        }
        dfs1(1, 1);
        dfs2(1, 1);
        dfs3(1, 1);
        build(1, 1, n);
        while(m--){
            int x = read(), y = read();
            int LCA = lca(x, y);
            printf("%d
", ask_he(x, y) - a[dfn[LCA]]);
        }
        puts(" ");
    }
    return 0; 
} 

但是啊，悄悄告诉你呀，这个方法，超慢哒！ 好奇怪的语气

根据ZHT大佬的思路，求出每个点到根的距离，然后每次询问直接为 $dis_x + dis_y - 2 * dis_{lca(x, y)}$

在此再次%%% + orz

树剖版：

#include <bits/stdc++.h>
using namespace std;
#define N 10010

inline int read(){
    int x = 0, s = 1;
    char c = getchar();
    while(!isdigit(c)){
        if(c == '-') s = -1;
        c = getchar();
    }
    while(isdigit(c)){
        x = x * 10 + (c ^ '0');
        c = getchar(); 
    }
    return x * s;
}

struct node{
    int v, w;
    int next;
} t[N << 1];
int f[N];

int bian = 0;
inline void add(int u, int v, int w){
    t[++bian] = (node){v, w, f[u]}, f[u] = bian;
    t[++bian] = (node){u, w, f[v]}, f[v] = bian;
    return ;
}  

int dfn[N], cnt = 0;
int fa[N], son[N], top[N], siz[N], deth[N];
int a[N];
int dis[N];
int n, m;

/*---------树链剖分start--------*/
#define v t[i].v
void dfs1(int now, int father){
    deth[now] = deth[father] + 1;
    siz[now] = 1;
    fa[now] = father;
    for(int i = f[now]; i; i = t[i].next){
        if(v != father){
            dis[v] = dis[now] + t[i].w;
            dfs1(v, now);
            siz[now] += siz[v];
            if(siz[v] > siz[son[now]])
                son[now] = v;
        }
    }
    return ;
}

void dfs2(int now, int tp){
    top[now] = tp;
    dfn[now] = ++cnt;
    if(!son[now]) return ;
    dfs2(son[now], tp);
    for(int i = f[now]; i; i = t[i].next){
        if(v != fa[now] && v != son[now])
            dfs2(v, v);
    }
    return ;
}

int LCA(int x, int y){
    while(top[x] != top[y]){
        if(deth[top[x]] < deth[top[y]]) swap(x, y);
        x = fa[top[x]];
    }
    return deth[x] < deth[y] ? x : y;
}

inline void clean(){
    memset(son, 0, sizeof(son));
    memset(dfn, 0, sizeof(dfn));
    cnt = 0;
    for(int i = 1;i <= bian; i++)
        t[i] = (node){0, 0, 0};
    memset(f, 0, sizeof(f));
    bian = 0;
    return ;
}

int main(){
    int T = read();
    while(T--){
        clean();
        int n = read(), m = read();
        for(int i = 1; i < n; i++){
            int x = read(), y = read(), w = read();
            add(x, y, w); 
        }
        dfs1(1, 1);
        dfs2(1, 1);
        while(m--){
            int x = read(), y = read();
            printf("%d
", dis[x] + dis[y] - 2 * dis[LCA(x, y)]);
        }
        puts(" ");
    }
}

 倍增版：

#include <bits/stdc++.h>
using namespace std;
#define N 10010

inline int read(){
    int x = 0, s = 1;
    char c = getchar();
    while(!isdigit(c)){
        if(c == '-') s = -1;
        c = getchar();
    }
    while(isdigit(c)){
        x = x * 10 + (c ^ '0');
        c = getchar(); 
    }
    return x * s;
}

struct node{
    int v, w;
    int next;
} t[N << 1];
int f[N];

int bian = 0;
inline void add(int u, int v, int w){
    t[++bian] = (node){v, w, f[u]}, f[u] = bian;
    t[++bian] = (node){u, w, f[v]}, f[v] = bian;
    return ;
}  

int a[N][21];
int fa[N], dis[N];
int deth[N];
int n, m;

#define v t[i].v

void dfs(int now, int father){
    a[now][0] = father;
    deth[now] = deth[father] + 1;
    for(int i = 1; (1 << i) <= deth[now]; i++)
        a[now][i] = a[a[now][i-1]][i-1];
    for(int i = f[now]; i; i = t[i].next){
        if(v != father){
            dis[v] = dis[now] + t[i].w;
            dfs(v, now);
        }
    }
    return ;
}

#undef v

int LCA(int x, int y){
    if(deth[x] < deth[y]) swap(x, y);
    for(int i = 20; i >= 0; i--){
        if(deth[a[x][i]] >= deth[y])
            x = a[x][i];
    }
    if(x == y) return x;
    for(int i = 20; i >= 0; i--){
        if(a[x][i] == a[y][i])
            continue;
        else x = a[x][i], y = a[y][i];
    }
    return a[x][0];
}

inline void clean(){
    memset(a, 0, sizeof(a));
    memset(deth, 0, sizeof(deth));
    memset(dis, 0, sizeof(dis));
    for(int i = 1;i <= bian; i++)
        t[i] = (node){0, 0, 0};
    memset(f, 0, sizeof(f));
    bian = 0;
    return ;
}

int main(){
    int T = read();
    while(T--){
        clean();
        int n = read(), m = read();
        for(int i = 1; i < n; i++){
            int x = read(), y = read(), w = read();
            add(x, y, w); 
        }
        dis[1] = 0;
        a[1][0] = 1;
        dfs(1, 1);
        while(m--){
            int x = read(), y = read();
            printf("%d
", dis[x] + dis[y] - dis[LCA(x, y)] * 2); 
        }
        puts(" ");
    }
    return 0;
}