Sending email - UVa 10986 Dijkstra+优先队列优化

最新推荐文章于 2020-03-20 17:34:14 发布

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本文介绍如何使用Dijkstra算法解决从S到T的最短路径问题，并提供了AC代码实现。

Sending email
Time Limit: 3 seconds

"A new internet watchdog is creating a stir in
Springfield. Mr. X, if that is his real name, has
come up with a sensational scoop."

Kent Brockman

There are n SMTP servers connected by network cables. Each of the m cables connects two computers and has a certain latency measured in milliseconds required to send an email message. What is the shortest time required to send a message from server S to server T along a sequence of cables? Assume that there is no delay incurred at any of the servers.

Input
The first line of input gives the number of cases, N. N test cases follow. Each one starts with a line containing n (2<=n<20000), m (0<=m<50000), S (0<=S<n) and T (0<=T<n). S!=T. The next m lines will each contain 3 integers: 2 different servers (in the range [0, n-1]) that are connected by a bidirectional cable and the latency, w, along this cable (0<=w<=10000).

Output
For each test case, output the line "Case #x:" followed by the number of milliseconds required to send a message from S to T. Print "unreachable" if there is no route from S to T.

Sample Input	Sample Output
3 2 1 0 1 0 1 100 3 3 2 0 0 1 100 0 2 200 1 2 50 2 0 0 1	Case #1: 100 Case #2: 150 Case #3: unreachable

题意：从S到T的最短路径。

思路：裸Dijkstra，不过因为数的范围太大，要用链表来存边，还要用优先队列优化。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
struct node
{
    int v,len;
    node *next;
}*Head[20010],tree[100010];
struct node2
{
    int val,pos;
    bool operator<(const node2 A)const
    {
        return A.val<val;
    }
};
int T,t,ptr,n,m,S,E,f[20010],INF=1e9,vis[20010];
priority_queue<node2> qu;
void AddEdge(int a,int b,int c)
{
    tree[ptr].v=b;
    tree[ptr].len=c;
    tree[ptr].next=Head[a];
    Head[a]=&tree[ptr++];
}
void Dijkstra()
{
    int i,j,k,ret,pos,v,len;
    for(i=0;i<n;i++)
       f[i]=INF;
    f[S]=0;
    node *p;
    node2 A;
    while(!qu.empty())
      qu.pop();
    A.val=0;
    A.pos=S;
    qu.push(A);
    for(i=1;i<=n;i++)
    {
        ret=INF;
        while(true)
        {
            if(qu.empty())
              break;
            A=qu.top();
            if(vis[A.pos]==t)
              qu.pop();
            else
            {
                pos=A.pos;
                break;
            }
        }
        if(qu.empty())
          break;
        vis[pos]=t;
        p=Head[pos];
        while(p)
        {
            v=p->v;
            len=p->len;
            if(f[pos]+len<f[v])
            {
                A.val=f[pos]+len;
                A.pos=v;
                qu.push(A);
                f[v]=f[pos]+len;
            }
            p=p->next;
        }
    }
}
int main()
{
    int i,j,k,u,v,w;
    scanf("%d",&T);
    for(t=1;t<=T;t++)
    {
        scanf("%d%d%d%d",&n,&m,&S,&E);
        memset(Head,NULL,sizeof(Head));
        ptr=0;
        for(i=1;i<=m;i++)
        {
            scanf("%d%d%d",&u,&v,&w);
            AddEdge(u,v,w);
            AddEdge(v,u,w);
        }
        Dijkstra();
        if(f[E]>=INF)
          printf("Case #%d: unreachable\n",t);
        else
          printf("Case #%d: %d\n",t,f[E]);
    }
}