洛谷P2912[USACO08OCT]牧场散步Pasture Walking

Posted 2020-09-30 北爱荒凉

题目描述

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)头奶牛，编号为1到W，它们正在同样编号为1到N的牧场上行走.为了方 便，我们假设编号为i的牛恰好在第i号牧场上.

有一些牧场间每两个牧场用一条双向道路相连，道路总共有N - 1条，奶牛可以在这些道路 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.在任意两个牧场间，有且仅有一条由若干道路组成的路径相连.也就是说，所有的道路构成了一棵树.

奶牛们十分希望经常互相见面.它们十分着急，所以希望你帮助它们计划它们的行程，你只 需要计算出Q(1 < Q < 1000)对点之间的路径长度?每对点以一个询问p1,p2 (1 <= p1 <= N; 1 <= p2 <= N). 的形式给出.

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.

 

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 

输出样例#1：

2 
7 

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.

题解：

LCA问题,dis[x]表示x到根节点的距离,设x,y的最近公共祖先为z,则x,y之间的距离为dis[x]+dis[y]-2*dis[z]。

倍增：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1000+5;
inline int read()
{
    int x=0,f=1; char ch=getchar();
    while(ch<‘0‘||ch>‘9‘) {if(ch==‘-‘)f=-1; ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘; ch=getchar();}
    return x*f;
}
int n,m,num;
int head[maxn],f[maxn][20],dep[maxn],dis[maxn];
bool vis[maxn];
struct node
{
    int next,to,dist;
}e[maxn<<1];
inline void add(int from,int to,int dist)
{
    e[++num].next=head[from];
    e[num].to=to;
    e[num].dist=dist;
    head[from]=num;
}
inline void dfs(int x,int d)
{
    vis[x]=1;dep[x]=d;
    for(int i=head[x];i;i=e[i].next)
    {
        int to=e[i].to;
        if(!vis[to])
        {
            dis[to]=dis[x]+e[i].dist;
            f[to][0]=x;
            dfs(to,d+1);
        }
    }
}
inline int lca(int a,int b)
{
    if(dep[a]<dep[b]){int t=a;a=b;b=t;}
    int d=dep[a]-dep[b];
    for(int i=0;i<=10;i++)
    if(d&(1<<i)) a=f[a][i];
    if(a==b) return a;
    for(int i=10;i>=0;i--)
    if(f[a][i]!=f[b][i])
    {
        a=f[a][i];
        b=f[b][i];
    }
    return f[a][0];
}
int main()
{
    n=read();m=read();
    for(int i=1;i<n;i++)
    {
        int x,y,z;
        x=read();y=read();z=read();
        add(x,y,z);add(y,x,z);
    }
    dfs(1,1);
    for(int j=1;j<=10;j++)
    for(int i=1;i<=n;i++)
    f[i][j]=f[f[i][j-1]][j-1];
    for(int i=1;i<=m;i++)
    {
        int a,b;
        a=read();b=read();
        printf("%d\n",dis[a]+dis[b]-(dis[lca(a,b)]<<1));
    }
    return 0;
}

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tarjan算法：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1000+5;
inline int read()
{
    int x=0,f=1; char ch=getchar();
    while(ch<‘0‘||ch>‘9‘) {if(ch==‘-‘)f=-1; ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘; ch=getchar();}
    return x*f;
}
int n,m,num,qnum;
int head[maxn],qhead[maxn],father[maxn],a[maxn][3],dis[maxn];
bool vis[maxn];
struct node
{
    int next,to,v;
}e[maxn<<1],q[maxn<<1];
void add(int from,int to,int v)
{
    e[++num].next=head[from];
    e[num].to=to;
    e[num].v=v;
    head[from]=num;
}
void qadd(int from,int to,int v)
{
    q[++qnum].next=qhead[from];
    q[qnum].to=to;
    q[qnum].v=v;
    qhead[from]=qnum;
}
int find(int x)
{
    if(x!=father[x]) father[x]=find(father[x]);
    return father[x];
}
void merge(int x,int y)
{
    int r1=find(x);
    int r2=find(y);
    father[r1]=r2;
}
void tarjan(int x)
{
    vis[x]=1;
    for(int i=qhead[x];i;i=q[i].next)
    {
        int to=q[i].to,v=q[i].v;
        if(vis[to]) a[v][2]=find(to);
    }
    for(int i=head[x];i;i=e[i].next)
    {
        int to=e[i].to;
        if(!vis[to])
        {
            dis[to]=dis[x]+e[i].v;
            tarjan(to);
            merge(to,x);
        }
    }
}
int main()
{
    n=read();m=read();
    for(int i=1;i<=n;i++) father[i]=i; 
    for(int i=1;i<n;i++)
    {
        int x,y,z;
        x=read();y=read();z=read();
        add(x,y,z);add(y,x,z);
    }
    for(int i=1;i<=m;i++)
    {
        a[i][0]=read();a[i][1]=read();
        qadd(a[i][0],a[i][1],i);
        qadd(a[i][1],a[i][0],i);
    }
    dis[1]=0;
    tarjan(1);
    for(int i=1;i<=m;i++)
    printf("%d\n",dis[a[i][0]]+dis[a[i][1]]-(dis[a[i][2]]<<1));
    return 0;
}
    

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