Connections between cities

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8041    Accepted Submission(s): 1994

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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最短路;

#include <cstdio>
#include <cstring>

#define N 10010
#define M 20010
#define C 2000010

struct Edge{
    int to, next, dis;
    Edge() {}
    Edge(int to, int dis, int next): to(to), dis(dis), next(next){};
}E[M];

struct Edge2{
    int to, next, w;
    Edge2() {}
    Edge2(int to, int w, int next): to(to), w(w), next(next) {};
}E2[C];

int n, m, c, tot, tot2;
int head[N], head2[N], dis[N], f[N], vis[N];

void addEdge(int u, int v, int dis){
    E[tot]= Edge(v, dis, head[u]);
    head[u]=tot++;
    E[tot]= Edge(u, dis, head[v]);
    head[v]=tot++;
}

void addEdge2(int u, int v){
    E2[tot2]= Edge2(v, -1, head2[u]);
    head2[u]=tot2++;
    E2[tot2]= Edge2(u, -1, head2[v]);
    head2[v]=tot2++;
}
/*void addEdge(int u, int v, int dis) {
    E[tot].to = v; E[tot].next = head[u]; E[tot].dis = dis; head[u] = tot++;
    u = u ^ v; v = u ^ v; u = u ^ v;
    E[tot].to = v; E[tot].next = head[u]; E[tot].dis = dis; head[u] = tot++;
}

void addEdge2(int u, int v) {
    E2[tot2].to = v; E2[tot2].next = head2[u]; E2[tot2].w = -1; head2[u] = tot2++;
    u = u ^ v; v = u ^ v; u = u ^ v;
    E2[tot2].to = v; E2[tot2].next = head2[u]; E2[tot2].w = -1; head2[u] = tot2++;
}*/
void init()
{
    memset(head, -1, sizeof(head));
    memset(head2, -1, sizeof(head2));
    tot=tot2=0;

    int u, v, d;
    for(int i=0; i< m; i++)
    {
        scanf("%d%d%d", &u, &v, &d);
        addEdge(u, v, d);
    } 

    for(int i=0; i<c; i++)
    {
        scanf("%d%d", &u, &v);
        addEdge2(u, v);
    }
    memset(vis, 0, sizeof(vis));
}

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

void tarjan(int u, int time)
{
    vis[u]= time;
    f[u]= u;

    int v;
    for(int i=head[u]; ~i; i=E[i].next)
    {
        v=E[i].to;
        if(vis[v])
            continue;
        dis[v]= dis[u]+ E[i].dis;
        tarjan(v, time);
        f[v]= u;
    } 

    for(int i= head2[u]; ~i; i=E2[i].next)
    {
        v=E2[i].to;
        if(vis[v]== time)
            E2[i].w= E2[i^1].w = dis[u]+dis[v]- 2*dis[find(v)];
    }
}
void solve()
{
    int cnt=1;
    for(int i=1; i<=n; i++){
        if(!vis[i]){
            dis[i]= 0;
            tarjan(i, cnt);
        }
        cnt++;
    }

    for(int i=0; i< tot2; i +=2){
        if(E2[i].w == -1)
            printf("Not connected\n");
        else
            printf("%d\n", E2[i].w)  ;
    }
}
int main()
{
    while(scanf("%d%d%d", &n, &m, &c) != EOF)
    {
        init();
        solve();
    }
    return 0;
}