ZOJ3195 Design the city

Design the city

Time Limit: 1 Second Memory Limit: 32768 KB

Cerror is the mayor of city HangZhou. As you may know, the traffic system of this city is so terrible, that there are traffic jams everywhere. Now, Cerror finds out that the main reason of them is the poor design of the roads distribution, and he want to change this situation.

In order to achieve this project, he divide the city up to N regions which can be viewed as separate points. He thinks that the best design is the one that connect all region with shortest road, and he is asking you to check some of his designs.

Now, he gives you an acyclic graph representing his road design, you need to find out the shortest path to connect some group of three regions.

Input

The input contains multiple test cases! In each case, the first line contian a interger N (1 < N < 50000), indicating the number of regions, which are indexed from 0 to N-1. In each of the following N-1 lines, there are three interger Ai, Bi, Li (1 < Li < 100) indicating there's a road with length Li between region Ai and region Bi. Then an interger Q (1 < Q < 70000), the number of group of regions you need to check. Then in each of the following Q lines, there are three interger Xi, Yi, Zi, indicating the indices of the three regions to be checked.

Process to the end of file.

Output

Q lines for each test case. In each line output an interger indicating the minimum length of path to connect the three regions.

Output a blank line between each test cases.

Sample Input

4
0 1 1
0 2 1
0 3 1
2
1 2 3
0 1 2
5
0 1 1
0 2 1
1 3 1
1 4 1
2
0 1 2
1 0 3
Sample Output

3
2

2

2

题意：要求求三个点连通的距离。。。

lca模板题。。求三个点就是每两个点之间相加，然后将那三个加起来除以2就是答案了，就是dist[x1]+dist[x2]+dist[x3]-dist[lca[x1,x2]]-dist[lca[x1,x3]]-dist[lca[x2,x3]]。

dist[x1]+dist[x2]-2*dist[lca[x1,x2]].....dist[x2]+dist[x3]-2*dist[x2,x3].....dist[x1]+dist[x3]-2*dist[x1,x3]....这三个式子相加除以2，就是上面的式子。。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN=50010;
const int MAXE=140010;
struct EDGE
{
    int v,next;
    int dis;
}edge[MAXN<<1];
struct QUEST
{
    int v,next;
    int num;
}q[MAXE*3];
int head1[MAXN],head2[MAXN],size;
bool vis[MAXN];
int fa[MAXN],dist[MAXN],ans1[MAXE>>1],ans2[MAXE>>1],ans3[MAXE>>1],lca[MAXE*3];
void init()
{
    memset(head1,-1,sizeof(head1));
    memset(head2,-1,sizeof(head2));
    memset(dist,0,sizeof(dist));
    memset(vis,0,sizeof(vis));
    size=0;
}
void add_edge(int u,int v,int dis)
{
    edge[size].v=v;
    edge[size].dis=dis;
    edge[size].next=head1[u];
    head1[u]=size++;
}
void add_q(int u,int v,int i)
{
    q[size].v=v;
    q[size].num=i;
    q[size].next=head2[u];
    head2[u]=size++;
}
int find(int x)
{
    return fa[x]==x?x:fa[x]=find(fa[x]);
}
void tarjan(int u)
{
    fa[u]=u;
    vis[u]=1;
    for(int i=head2[u];i!=-1;i=q[i].next)
    {
        int v=q[i].v;
        if(vis[v])
        {
            lca[q[i].num]=find(v);
        }
    }
    for(int i=head1[u];i!=-1;i=edge[i].next)
    {
        int v=edge[i].v;
        if(!vis[v])
        {
            dist[v]=dist[u]+edge[i].dis;
            tarjan(v);
            fa[v]=u;
        }
    }
}
int main()
{
    int n,m,i,flag=1;
    //freopen("in.txt","r",stdin);
    while(scanf("%d",&n)==1)
    {
        if(!flag)
            printf("\n");
        flag=0;
        init();
        int u,v,c;
        for(i=0;i<n-1;i++)
        {
            scanf("%d%d%d",&u,&v,&c);
            add_edge(u,v,c);
            add_edge(v,u,c);
        }
        scanf("%d",&m);
        int x1,x2,x3;
        memset(lca,0,sizeof(lca));
        size=0;
        for(i=0;i<m;i++)
        {
            scanf("%d%d%d",&x1,&x2,&x3);
            add_q(x1,x2,i);
            add_q(x2,x1,i);
            add_q(x1,x3,i+m);
            add_q(x3,x1,i+m);
            add_q(x2,x3,i+2*m);
            add_q(x3,x2,i+2*m);
            ans1[i]=x1;
            ans2[i]=x2;
            ans3[i]=x3;
        }
        tarjan(1);
        for(i=0;i<m;i++)
        {
            printf("%d\n",dist[ans1[i]]+dist[ans2[i]]+dist[ans3[i]]-dist[lca[i]]-dist[lca[i+m]]-dist[lca[i+2*m]]);
        }
    }
    return 0;
}