POJ 1986

题目链接：http://poj.org/problem?id=1986

Description

Farmer John's cows refused to run in his marathon since he chose a path much too long for their leisurely lifestyle. He therefore wants to find a path of a more reasonable length. The input to this problem consists of the same input as in "Navigation Nightmare",followed by a line containing a single integer K, followed by K "distance queries". Each distance query is a line of input containing two integers, giving the numbers of two farms between which FJ is interested in computing distance (measured in the length of the roads along the path between the two farms). Please answer FJ's distance queries as quickly as possible! 

Input

* Lines 1..1+M: Same format as "Navigation Nightmare" 

* Line 2+M: A single integer, K. 1 <= K <= 10,000 

* Lines 3+M..2+M+K: Each line corresponds to a distance query and contains the indices of two farms. 

Output

* Lines 1..K: For each distance query, output on a single line an integer giving the appropriate distance. 

Sample Input

7 6
1 6 13 E
6 3 9 E
3 5 7 S
4 1 3 N
2 4 20 W
4 7 2 S
3
1 6
1 4
2 6

Sample Output

13
3
36

Hint

Farms 2 and 6 are 20+3+13=36 apart. 

 

题意：

输入第1~M+1行，与POJ 1984相同，代表了农场地图。

然后再一行有一个整数K代表询问数，

再然后有K个询问u和v之间最短距离。

 

题解：

本题的输入确定了农场地图是一棵树，并且本题不需要知道农场之间的位置关系，所以不需要记录东西南北。

树上两点间的最短距离，有两种情况：

①u是v的祖先，则dist(u,v) = dist(root,v) - dist(root,u)

②u不是v的祖先，那么从u到v必然要经过LCA(u,v)，显然就是最短路径，则dist(u,v) = dist(root,u) - dist(root,LCA(u,v)) + dist(root,v) - dist(root,LCA(u,v))

不难发现，第①种情况下，dist(root,LCA(u,v)) = dist(root,u)，那么①和②就可以统一为：dist(u,v) = dist(root,u) + dist(root,v) - 2 * dist(root,LCA(u,v))

所以我们只要计算出每个节点和树根的距离，求出所有查询(u,v)的LCA(u,v)，就能得到dist(u,v)。

 

AC代码：

#include<cstdio>
#include<vector>
using namespace std;

const int maxn=40000+5; //节点数
const int maxm=40000+5; //边数
const int maxq=10000+5; //查询数

int par[maxn];
int find(int x){return (par[x]==x)?x:(par[x]=find(par[x]));}

struct Edge{
    int u,v,w;
    Edge(int u=0,int v=0,int w=0){this->u=u,this->v=v,this->w=w;}
};
vector<Edge> E;
vector<int> Ge[maxn];
void addedge(int u,int v,int w)
{
    E.push_back(Edge(u,v,w));
    Ge[u].push_back(E.size()-1);
}

struct Query{
    int u,v;
    int lca;
    Query(int u=0,int v=0,int lca=0){this->u=u,this->v=v,this->lca=lca;}
};
vector<Query> Q;
vector<int> Gq[maxn];
void addquery(int u,int v)
{
    Q.push_back(Query(u,v));
    Gq[u].push_back(Q.size()-1);
}

bool vis[maxn];
int dist[maxn];
void LCA(int u,int d)
{
    par[u]=u; //建立以u为代表元素的集合
    vis[u]=1;
    dist[u]=d;
    for(int i=0;i<Ge[u].size();i++)
    {
        Edge &e=E[Ge[u][i]]; int v=e.v;
        if(!vis[v])
        {
            LCA(v,d+e.w);
            par[v]=u; //将v的集合并入u的集合
        }
    }
    for(int i=0;i<Gq[u].size();i++)
    {
        Query &q=Q[Gq[u][i]]; int v=q.v;
        if(vis[v])
        {
            q.lca=find(v);
            Q[Gq[u][i]^1].lca=q.lca;
        }
    }
}

int m,n,k;
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++)
    {
        int u,v,w; char d[3];
        scanf("%d%d%d%s",&u,&v,&w,d);
        addedge(u,v,w);
        addedge(v,u,w);
    }

    scanf("%d",&k);
    for(int i=1;i<=k;i++)
    {
        int u,v;
        scanf("%d%d",&u,&v);
        addquery(u,v);
        addquery(v,u);
    }

    LCA(1,0);

    for(int i=1;i<=k;i++)
    {
        printf("%d
",dist[Q[(i-1)*2].u]+dist[Q[(i-1)*2].v]-2*dist[Q[(i-1)*2].lca]);
    }
}