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【洛谷P2912】 [USACO08OCT]牧场散步Pasture Wal…

2018-06-17 22:14:19来源：未知阅读 ()

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N，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：

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：

Tarjan:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1000+5;
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,num1,num2;
int head1[maxn],head2[maxn],a[maxn][3],dis[maxn],father[maxn];
bool vis[maxn];
struct node
{
    int next,to,v;
}e[maxn*2],q[maxn*2];
void add1(int from,int to,int dist)
{
    e[++num1].next=head1[from];
    e[num1].to=to;
    e[num1].v=dist;
    head1[from]=num1;
}
void add2(int from,int to,int id)
{
    q[++num2].next=head2[from];
    q[num2].to=to;
    q[num2].v=id;
    head2[from]=num2;
}
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=head2[x];i;i=q[i].next)
    {
        int to=q[i].to,id=q[i].v;
        if(vis[to]) a[id][2]=find(to);
    }
    for(int i=head1[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();
        add1(x,y,z);add1(y,x,z);
    }
    for(int i=1;i<=m;i++)
    {
        a[i][0]=read();a[i][1]=read();
        add2(a[i][0],a[i][1],i);
        add2(a[i][1],a[i][0],i);
    }
    tarjan(1);
    for(int i=1;i<=m;i++)
    printf("%d\n",dis[a[i][0]]+dis[a[i][1]]-2*dis[a[i][2]]);
    return 0;
}

倍增:

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int n,q,cnt;
int deep[1001],head[1001],dis[1001],fa[1001][11];
bool vis[1001];
struct data{int to,next,v;}e[2001];
void ins(int u,int v,int w)
{e[++cnt].to=v;e[cnt].next=head[u];e[cnt].v=w;head[u]=cnt;}
void insert(int u,int v,int w)
{ins(u,v,w);ins(v,u,w);}
void dfs(int x)
{
    vis[x]=1;
    for(int i=1;i<=9;i++)
    {
        if(deep[x]<(1<<i))break;
        fa[x][i]=fa[fa[x][i-1]][i-1];
    }
    for(int i=head[x];i;i=e[i].next)
    {
        if(vis[e[i].to])continue;
        deep[e[i].to]=deep[x]+1;
        dis[e[i].to]=dis[x]+e[i].v;
        fa[e[i].to][0]=x;
        dfs(e[i].to);
    }
}
int lca(int x,int y)
{
    if(deep[x]<deep[y])swap(x,y);
    int d=deep[x]-deep[y];
    for(int i=0;i<=9;i++)
       if((1<<i)&d)x=fa[x][i];
    for(int i=9;i>=0;i--)
       if(fa[x][i]!=fa[y][i]) 
          {x=fa[x][i];y=fa[y][i];}
    if(x==y)return x;
    else return fa[x][0];
}
int main()
{
    scanf("%d%d",&n,&q);
    for(int i=1;i<n;i++)
    {
        int u,v,w;
        scanf("%d%d%d",&u,&v,&w);
        insert(u,v,w);
    }
    dfs(1);
    for(int i=1;i<=q;i++)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        printf("%d\n",dis[x]+dis[y]-2*dis[lca(x,y)]);
    }
    return 0;
}