洛谷 P2912 [USACO08OCT]牧场走走Pasture Walking

P2912 [USACO08OCT]牧场走走Pasture Walking,p2912pasture

说明

Query 1: The walkway between pastures 1 and 2 has length 2.

Query 2: Travel through the walkway between pastures 3 and 4, then the
one between 4 and 1, and finally the one between 1 and 2, for a total
length of 7.

 

裸LCA,预管理的时候顺便求到根节点的路线长,答案为len[u] + len[v] – 2
* len[lca(va, vb)]

 

 1 #include <bits/stdc++.h>
 2 
 3 const int MAXN = 5000 + 10;
 4 
 5 inline void read(int &x)
 6 {
 7     x = 0;char ch = getchar();char c = ch;
 8     while(ch > '9' || ch < '0')c = ch, ch = getchar();
 9     while(ch <= '9' && ch >= '0')x = x * 10 + ch - '0', ch = getchar();
10     if(c == '-')x = -x;
11 }
12 
13 inline void swap(int &a, int &b)
14 {
15     int tmp = a;
16     a = b;
17     b = tmp;
18 }
19 
20 struct Edge{int u,v,w,next;}edge[MAXN >> 1];
21 int head[MAXN],root,cnt,deep[MAXN],len[MAXN],p[MAXN][20];
22 int n,m,tmp1,tmp2,tmp3;
23 bool b[MAXN];
24 
25 inline void insert(int a, int b, int c)
26 {
27     edge[++cnt] = {a,b,c,head[a]};
28     head[a] = cnt;
29 }
30 
31 void dfs(int u, int step)
32 {
33     for(int pos = head[u];pos;pos = edge[pos].next)
34     {
35         int v = edge[pos].v;
36         if(!b[v])
37         {
38             b[v] = true;
39             deep[v] = step + 1;
40             len[v] = len[u] + edge[pos].w;
41             p[v][0] = u;
42             dfs(v, step + 1);
43         }
44     }
45 }
46 
47 inline void yuchuli()
48 {
49     b[root] = true;
50     len[root] = 0;
51     deep[root] = 0;
52     dfs(root, 0);
53     for(int i = 1;(1 << i) <= n;i ++)
54     {
55         for(int j = 1;j <= n;j ++)
56         {
57             p[j][i] = p[p[j][i - 1]][i - 1];
58         }
59     }
60 }
61 
62 inline int lca(int va, int vb)
63 {
64     if(deep[va] < deep[vb])swap(va, vb);
65     int M = 0;
66     for(;(1 << M) <= n;M ++);
67     M --;
68     for(int i = M;i >= 0;i --)
69         if(deep[va] - (1 << i) >= deep[vb])
70             va = p[va][i];
71     if(va == vb)return va;
72     for(int i = M;i >= 0;i --)
73         if(p[va][i] != p[vb][i])
74         {
75             va = p[va][i];
76             vb = p[vb][i];
77         }
78     return p[va][0];
79 }
80 
81 int main()
82 {
83     read(n);read(m);
84     for(int i = 1;i < n;i ++)
85     {
86         read(tmp1);read(tmp2);read(tmp3);
87         insert(tmp1, tmp2, tmp3);
88         insert(tmp2, tmp1, tmp3);
89     }
90     root = 1;
91     yuchuli();
92     for(int i = 1;i <= m;i ++)
93     {
94         read(tmp1);read(tmp2);
95         printf("%d\n", len[tmp1] + len[tmp2] - 2 * len[lca(tmp1, tmp2)]);
96     }
97     return 0;
98 }

 

 

标题陈诉

The N cows (2 <= N <= 1,000) conveniently numbered 1..N are
grazing among the N pastures also conveniently numbered 1..N. Most
conveniently of all, cow i is grazing in pasture i.

Some pairs of pastures are connected by one of N-1 bidirectional
walkways that the cows can traverse. Walkway i connects pastures A_i
and B_i (1 <= A_i <= N; 1 <= B_i <= N) and has a length
of L_i (1 <= L_i <= 10,000).

The walkways are set up in such a way that between any two distinct
pastures, there is exactly one path of walkways that travels between
them. Thus, the walkways form a tree.

The cows are very social and wish to visit each other often. Ever in a
hurry, they want you to help them schedule their visits by computing the
lengths of the paths between 1 <= L_i <= 10,000 pairs of pastures
(each pair given as a query p1,p2 (1 <= p1 <= N; 1 <= p2 <=
N).

POINTS: 200

有N(2<=N<=1000)头白牛,编号为1到N,它们正在同样号码为1到N的牧场上行走.为了方
便,大家假诺编号为i的牛恰辛亏第i号牧场上.

有一部分牧场间每四个牧场用一条双向道路相连,道路总共有N –
1条,红牛能够在这几个道路
上行走.第i条道路把第Ai个牧场和第Bi个牧场连了起来(1 <= A_i <= N; 1
<= B_i <= N),而它的长度 是 1 <= L_i <=
10,000.在自便多个牧场间,有且只有一条由若干道路组成的路线相连.也即是说,全部的道路构成了一棵树.

红牛们极度企盼日常互相会见.它们非常焦急,所以希望你帮忙它们安排它们的行程,你只
须求总结出Q(1 < Q < 一千)对点之间的渠道长度•每对点以三个打探p1,p2
(1 <= p1 <= N; 1 <= p2 <= N). 的花样给出.

主题材料陈说

The N cows (2 <= N <= 1,000) conveniently numbered 1..N are
grazing among the N pastures also conveniently numbered 1..N. Most
conveniently of all, cow i is grazing in pasture i.

Some pairs of pastures are connected by one of N-1 bidirectional
walkways that the cows can traverse. Walkway i connects pastures A_i
and B_i (1 <= A_i <= N; 1 <= B_i <= N) and has a length
of L_i (1 <= L_i <= 10,000).

The walkways are set up in such a way that between any two distinct
pastures, there is exactly one path of walkways that travels between
them. Thus, the walkways form a tree.

The cows are very social and wish to visit each other often. Ever in a
hurry, they want you to help them schedule their visits by computing the
lengths of the paths between 1 <= L_i <= 10,000 pairs of pastures
(each pair given as a query p1,p2 (1 <= p1 <= N; 1 <= p2 <=
N).

POINTS: 200

有N(2<=N<=一千)头奶牛,编号为1到W,它们正在同样号码为1到N的牧场上行走.为了方
便,大家倘若编号为i的牛恰幸而第i号牧场上.

有部分牧场间每八个牧场用一条双向道路相连,道路总共有N –
1条,红牛能够在那一个道路
上行走.第i条道路把第Ai个牧场和第Bi个牧场连了起来(1 <= A_i <= N; 1
<= B_i <= N),而它的长短 是 1 <= L_i <=
10,000.在自由三个牧场间,有且唯有一条由若干道路组成的不二等秘书技相连.也正是说,全体的征程构成了一棵树.

水牛们极度意在平日相互会面.它们非常匆忙,所以希望你帮助它们布署它们的路途,你只
必要总括出Q(1 < Q < 一千)对点之间的路径长度•每对点以一个叩问p1,p2
(1 <= p1 <= N; 1 <= p2 <= N). 的样式给出.

说明

Query 1: The walkway between pastures 1 and 2 has length 2.

Query 2: Travel through the walkway between pastures 3 and 4, then the
one between 4 and 1, and finally the one between 1 and 2, for a total
length of 7.

输入输出样例

输入样例#1:

4 2 
2 1 2 
4 3 2 
1 4 3 
1 2 
3 2 

出口样例#1:

2 
7 

输入输出格式

输入格式:

  • Line 1: Two space-separated integers: N and Q

  • Lines 2..N: Line i+1 contains three space-separated integers: A_i,
    B_i, and L_i

  • Lines N+1..N+Q: Each line contains two space-separated integers
    representing two distinct pastures between which the cows wish to
    travel: p1 and p2

输出格式:

  • Lines 1..Q: Line i contains the length of the path between the two
    pastures in query i.

说明

Query 1: The walkway between pastures 1 and 2 has length 2.

Query 2: Travel through the walkway between pastures 3 and 4, then the
one between 4 and 1, and finally the one between 1 and 2, for a total
length of 7.

 

LCA裸题 

屠龙宝刀点击就送

#include <ctype.h>
#include <cstdio>
#define N 1005 

void read(int &x)
{
    x=0;register char ch=getchar();
    for(;!isdigit(ch);ch=getchar());
    for(;isdigit(ch);ch=getchar()) x=x*10+ch-'0';
}
struct Edge
{
    int next,to,dis;
    Edge (int next=0,int to=0,int dis=0) :next(next),to(to),dis(dis){}
}edge[N<<1];
int dis[N],dad[N][25],dep[N],head[N],cnt,n,q;
void insert(int u,int v,int w)
{
    edge[++cnt]=Edge(head[u],v,w);
    head[u]=cnt;
}
void swap(int &x,int &y)
{
    int tmp=y;
    y=x;
    x=tmp;
}
void dfs(int x)
{
    dep[x]=dep[dad[x][0]]+1;
    for(int i=0;dad[x][i];i++)
    dad[x][i+1]=dad[dad[x][i]][i];
    for(int u=head[x];u;u=edge[u].next)
    {
        int v=edge[u].to;
        if(dad[x][0]!=v)
        {
            dad[v][0]=x;
            dis[v]=dis[x]+edge[u].dis;
            dfs(v);
        }
    }
}
int lca(int x,int y)
{
    if(dep[x]>dep[y]) swap(x,y);
    for(int i=20;i>=0;i--)
    if(dep[dad[y][i]]>=dep[x]) y=dad[y][i];
    if(x==y) return x;
    for(int i=20;i>=0;i--)
    if(dad[y][i]!=dad[x][i]) y=dad[y][i],x=dad[x][i];
    return dad[x][0];
}
int main()
{
    read(n);
    read(q);
    for(int x,y,z,i=1;i<n;i++)
    {
        read(x);
        read(y);
        read(z);
        insert(x,y,z);
        insert(y,x,z);
    }
    dfs(1);
    for(int x,y;q--;)
    {
        read(x);
        read(y);
        int LCA=lca(x,y);
        printf("%d\n",dis[x]+dis[y]-2*dis[LCA]);
    }
    return 0;
}

 

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