hdu——2586 How far away ？

深碍√TFBOYSˉ_ 2021-09-20 07:14 285阅读 0赞

# How far away ？ #

**Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)  
Total Submission(s): 17449    Accepted Submission(s): 6747**

Problem Description

There are n houses in the village and some bidirectional roads connecting them. Every day peole always like to ask like this "How far is it if I want to go from house A to house B"? Usually it hard to answer. But luckily int this village the answer is always unique, since the roads are built in the way that there is a unique simple path("simple" means you can't visit a place twice) between every two houses. Yout task is to answer all these curious people.

Input

First line is a single integer T(T<=10), indicating the number of test cases.  
  For each test case,in the first line there are two numbers n(2<=n<=40000) and m (1<=m<=200),the number of houses and the number of queries. The following n-1 lines each consisting three numbers i,j,k, separated bu a single space, meaning that there is a road connecting house i and house j,with length k(0<k<=40000).The houses are labeled from 1 to n.  
  Next m lines each has distinct integers i and j, you areato answer the distance between house i and house j.

Output

For each test case,output m lines. Each line represents the answer of the query. Output a bland line after each test case.

Sample Input

2 3 2 1 2 10 3 1 15 1 2 2 3 2 2 1 2 100 1 2 2 1

Sample Output

10 25 100 100

Source

[ECJTU 2009 Spring Contest][]

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lcy   |   We have carefully selected several similar problems for you:   [3486][]  [2874][]  [2888][]  [3234][]  [2818][]

挨着做的，然后、、、、、

和上一题一样的水题、、、（模板题、、、）

代码：

#include<cstdio>
    #include<cstdlib>
    #include<cstring>
    #include<iostream>
    #include<algorithm>
    #define N 41000
    using namespace std;
    int t,n,m,x,y,z,ans,tot;
    int fa[N],top[N],size[N],deep[N],head[N],dis[N];
    struct Edge
    {
        int to,dis,from,next;
    }edge[N<<1];
    int add(int x,int y,int z)
    {
        tot++;
        edge[tot].to=y;
        edge[tot].dis=z;
        edge[tot].next=head[x];
        head[x]=tot;
    }
    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 begin()
    {
        ans=0;tot=0;
        memset(fa,0,sizeof(fa));
        memset(top,0,sizeof(top));
        memset(dis,0,sizeof(dis));
        memset(edge,0,sizeof(edge));
        memset(deep,0,sizeof(deep));
        memset(size,0,sizeof(size));
        memset(head,0,sizeof(head));
    }
    int dfs(int x)
    {
        size[x]=1;
        deep[x]=deep[fa[x]]+1;
        for(int i=head[x];i;i=edge[i].next)
        {
            int to=edge[i].to;
            if(fa[x]==to) continue;
            dis[to]=dis[x]+edge[i].dis;
            fa[to]=x,dfs(to);
            size[x]+=size[to];
        }
    }
    int dfs1(int x)
    {
        int t=0;
        if(!top[x]) top[x]=x;
        for(int i=head[x];i;i=edge[i].next)
        {
            int to=edge[i].to;
            if(fa[x]!=to&&size[t]<size[to]) t=to;
        }
        if(t) top[t]=top[x],dfs1(t);
        for(int i=head[x];i;i=edge[i].next)
        {
            int to=edge[i].to;
            if(fa[x]!=to&&t!=to)
             dfs1(to);
        }
    }
    int lca(int x,int y)
    {
        for(;top[x]!=top[y];x=fa[top[x]])
          if(deep[top[x]]<deep[top[y]]) swap(x,y);
        if(deep[x]>deep[y]) swap(x,y);
        return x;
    }
    int main()
    {
        t=read();
        while(t--)
        {
            n=read();m=read();begin();
            for(int i=1;i<n;i++)
            {
                x=read(),y=read(),z=read();
                add(x,y,z),add(y,x,z);
            }
            dfs(1),dfs1(1);
            for(int i=1;i<=m;i++)
            {
                x=read(),y=read();
                ans=dis[x]+dis[y]-2*dis[lca(x,y)];
                printf("%d\n",ans);
            }
        }
        return 0;
    }

转载于:https://www.cnblogs.com/z360/p/7412779.html

[ECJTU 2009 Spring Contest]: http://acm.hdu.edu.cn/search.php?field=problem&key=ECJTU+2009+Spring+Contest+&source=1&searchmode=source
[3486]: http://acm.hdu.edu.cn/showproblem.php?pid=3486
[2874]: http://acm.hdu.edu.cn/showproblem.php?pid=2874
[2888]: http://acm.hdu.edu.cn/showproblem.php?pid=2888
[3234]: http://acm.hdu.edu.cn/showproblem.php?pid=3234
[2818]: http://acm.hdu.edu.cn/showproblem.php?pid=2818