poj 2378 (dijkstra)

http://poj.org/problem?id=2387

一个dijkstra的模板题

#include <stdio.h>
#include <string.h>
#define inf 999999;

int graph[ 1005 ][ 1005 ];
bool mark[ 1005 ];
int m,n;

void dijkstra()
{
    int d[ 1005 ],k,Min;
    bool mark[ 1005 ];
    for(int i = 1 ; i <= n ; i++)       //对d[i]进行初始化
    {
        mark[ i ] = false;
        d[ i ] = graph[ 1 ][ i ];
    }
    for(int i = 1 ; i <= n ; i++)   
    {
        Min=inf;
        for(int j = 1 ; j <= n ; j++)
        {
            if(!mark[ j ] && d[ j ] < Min)    
            {
                k = j;
                Min = d[ j ];
            }
        }
        mark[ k ] = true;

        for(int j = 1 ; j <= n ; j++)
        {
            if(!mark[ j ] && d[ j ] > graph[ k ][ j ] + d[ k ])
                d[ j ] = graph[ k ] [ j ] + d[ k ];
        }
    }
    printf("%d\n",d[n]);
}

int main()
{
    int a,b,c;
    while(~scanf("%d%d",&m,&n)){
        for(int i = 0 ; i <= n ; i++)
            for(int j = 0 ; j <= n ; j++)
                if(i == j)
                    graph[ i ][ j ] = 0;
                else
                {
                    graph[ i ][ j ] = inf;
                    graph[ j ][ i ] = inf;
                }
            for(int i = 1 ; i <= m ;i++)
            {
                scanf("%d%d%d",&a,&b,&c);
                if(graph[ a ][ b ] > c)
                {
                   graph[ a ][ b ] = c;
                   graph[ b ][ a ] = c;
                }
            }
            dijkstra();
    }
    return 0;
}

 

 

 

dijkstra +head

 1 const int N=405;
 2 struct rec
 3 {
 4     int v,w;//v代表所连的值，W代表权值
 5 };
 6 
 7 
 8 vector<rec> edge[N*N];
 9 int n,st,ed;
10 long long dis[N*N];
11 bool vis[N*N];
12 
13 struct cmp
14 {
15     bool operator()(int a,int b)
16     {
17         return dis[a]>dis[b];
18     }
19 };
20 
21 void Dijkstra()
22 {
23     priority_queue<int,vector<int>,cmp> Q;
24     memset(dis,-1,sizeof(dis));
25     memset(vis,0,sizeof(vis));
26     int i,u,v;
27     Q.push(st);
28     dis[st]=1;
29     while(!Q.empty())
30     {
31         u=Q.top();
32         Q.pop();
33         vis[u]=0;
34         if(u==st)
35             break;
36             if(dis[v]==-1 || dis[v]<dis[u]*edge[u][i].w)
37             {
38                 dis[v]=dis[u]*edge[u][i].w;
39                 if(!vis[v])
40                 {
41                     vis[v]=1;
42                     Q.push(v);
43                 }
44             }
45         }
46     }
47 }
48         for(i=0; i<edge[u].size(); i++)
49         {
50             v=edge[u][i].v;
51             if(dis[v]==-1 || dis[v]<dis[u]*edge[u][i].w)
52             {
53                 dis[v]=dis[u]*edge[u][i].w;
54                 if(!vis[v])
55                 {
56                     vis[v]=1;
57                     Q.push(v);
58                 }
59             }
60         }
61     }
62 }

 

posted @ 2016-08-02 16:55  一个_小菜鸟  阅读(196)  评论(0)  编辑  收藏  举报