Connections between cities （hdu 2874 LCA）

Connections between cities

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 6908 Accepted Submission(s): 1798



Problem Description

After World War X, a lot of cities have been seriously damaged, and we need to rebuild those cities. However, some materials needed can only be produced in certain places. So we need to transport these materials from city to city. For most of roads had been
totally destroyed during the war, there might be no path between two cities, no circle exists as well.

Now, your task comes. After giving you the condition of the roads, we want to know if there exists a path between any two cities. If the answer is yes, output the shortest path between them.



Input

Input consists of multiple problem instances.For each instance, first line contains three integers n, m and c, 2<=n<=10000, 0<=m<10000, 1<=c<=1000000. n represents the number of cities numbered from 1 to n. Following m lines, each line has three integers i,
j and k, represent a road between city i and city j, with length k. Last c lines, two integers i, j each line, indicates a query of city i and city j.



Output

For each problem instance, one line for each query. If no path between two cities, output “Not connected”, otherwise output the length of the shortest path between them.



Sample Input

5 3 2
1 3 2
2 4 3
5 2 3
1 4
4 5

Sample Output

Not connected
6

Hint
Hint

Huge input, scanf recommended.

Source

2009 Multi-University Training Contest 8 - Host
by BJNU



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题意：n个点m条边c次询问，每次询问（u，v）之间的最短距离，若不联通，输出“Not connected”，否则输出最短距离，没有环。

思路：LCA，但是给的图是一个森林，每次找到子树Tarjan，还有一种方法就是添加根root，把所有树连成一棵树再一次LCA。数组开小了它却返回TLE，郁闷我半天。。。

代码：

#include <iostream>
#include <functional>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b)  for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b)  for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define DBG         pf("Hi\n")
typedef long long ll;
using namespace std;

#define INF 0x3f3f3f3f
#define mod 1000000009
const int maxn = 10005;
const int MAXN = 200005;
const int MAXM = 200010;
const int N = 1005;

struct Edge
{
    int u,v,w,next;
}edge[MAXN];

struct Query
{
    int q,id,next;
}query[2000000+10];

int n,m,c;
int head[maxn],hed[maxn];
int father[maxn],dis[maxn];
int tot1,tot2;
bool vis[maxn],vis2[maxn];
int ans[1000000+10];

void init()
{
    tot1=tot2=0;
    memset(head,-1,sizeof(head));
    memset(hed,-1,sizeof(hed));
    memset(vis2,false,sizeof(vis2));
    memset(dis,0,sizeof(dis));
    memset(ans,-1,sizeof(ans));
    for (int i=0;i<=n;i++)
        father[i]=i;
}

void addedge(int u,int v,int w)
{
    edge[tot1].u=u;
    edge[tot1].v=v;
    edge[tot1].w=w;
    edge[tot1].next=head[u];
    head[u]=tot1++;
}

void addQuery(int u,int v,int id)
{
    query[tot2].q=v;
    query[tot2].id=id;
    query[tot2].next=hed[u];
    hed[u]=tot2++;
}

int find_father(int x)
{
    if (x!=father[x])
        father[x]=find_father(father[x]);
    return father[x];
}

void Union(int u,int v)
{
    father[find_father(v)]=find_father(u);
}

void Tarjan(int u,int d)
{
    vis[u]=true;
    vis2[u]=true;
    dis[u]=d;
    for (int i=head[u];~i;i=edge[i].next)
    {
        int v=edge[i].v;
        if (vis[v]) continue;
        Tarjan(v,d+edge[i].w);
        Union(u,v);
    }
    for (int i=hed[u];~i;i=query[i].next)
    {
        int v=query[i].q;
        if (!vis[v]) continue;
        ans[query[i].id]=dis[u]+dis[v]-2*dis[find_father(v)];
    }
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("C:/Users/lyf/Desktop/IN.txt","r",stdin);
#endif
    int i,j,u,v,w;
    while (~scanf("%d%d%d",&n,&m,&c))
    {
        init();
        for (i=0;i<m;i++)
        {
            scanf("%d%d%d",&u,&v,&w);
            vis2[v]=true;
            addedge(u,v,w);
            addedge(v,u,w);
        }
        for (i=0;i<c;i++)
        {
            scanf("%d%d",&u,&v);
            addQuery(u,v,i);
            addQuery(v,u,i);
        }
        for (i=1;i<=n;i++)
        {
            if (!vis2[i])
            {
                memset(vis,false,sizeof(vis));
                Tarjan(i,0);
            }
        }
        for (i=0;i<c;i++)
        {
            if (ans[i]==-1) printf("Not connected\n");
            else printf("%d\n",ans[i]);
        }
    }
    return 0;
}