hdu 1690 构图后Floyd 数据很大

WA了好多次... 这题要用long long 而且INF要设大一点

Sample Input
2 //T
1 2 3 4 1 3 5 7 //L1-L4 C1-C4 距离和花费
4 2 //结点数 询问次数
1 //结点的横坐标
2
3
4
1 4 //起点 终点
4 1
1 2 3 4 1 3 5 7
4 1
1
2
3
10
1 4

Sample Output
Case 1:
The minimum cost between station 1 and station 4 is 3.
The minimum cost between station 4 and station 1 is 3.
Case 2:
Station 1 and station 4 are not attainable.

 

 1 # include <iostream>
 2 # include <cstdio>
 3 # include <cstring>
 4 # include <algorithm>
 5 # include <cmath>
 6 # include <queue>
 7 # define LL long long
 8 using namespace std ;
 9 
10 const LL INF=0x7f7f7f7f7f7f7f7fLL;
11 const int MAXN=210;
12 
13 LL L[10] ;
14 LL C[10] ;
15 LL x[MAXN] ;
16 
17 LL dis[MAXN][MAXN];
18 int n ;
19 
20 
21 void floyed()//节点从1~n编号
22 {
23     int i,j,k;
24     for(k=1;k<=n;k++)
25        for(i=1;i<=n;i++)
26          for(j=1;j<=n;j++)
27              if(dis[i][k]+dis[k][j] < dis[i][j] && dis[i][k] != INF && dis[k][j] != INF)
28                  dis[i][j]=dis[i][k]+dis[k][j];
29 
30 }
31 
32 LL Cost(LL d)
33 {
34     if (d < 0)
35         d *= -1 ;
36     if (d > 0 && d<= L[1])
37         return C[1] ;
38     if (d > L[1] && d<= L[2])
39         return C[2] ;
40     if (d > L[2] && d<= L[3])
41         return C[3] ;
42     if (d > L[3] && d<= L[4])
43         return C[4] ;
44     return INF ;
45 }
46 
47 int main ()
48 {
49    // freopen("in.txt","r",stdin) ;
50     int cnt ;
51     int T ;
52     cin>>T ;
53     int Case = 0 ;
54     while (T--)
55     {
56         Case++ ;
57         cout<<"Case "<<Case<<":"<<endl ;
58         int i , j  ;
59         LL w ;
60         for (i = 1 ; i <= 4 ; i++)
61             cin>>L[i];
62         for (i = 1 ; i <= 4 ; i++)
63             cin>>C[i];
64 
65         cin>>n>>cnt ;
66         for (i = 1 ; i <= n ; i++)
67             cin>>x[i];
68         for (i = 1 ; i <= n ; i++)
69             for (j = 1 ; j <= n ; j++)
70           {
71               if(i==j)dis[i][j]=0;
72               else dis[i][j]=INF;
73           }
74         for (i = 1 ; i <= n ; i++)
75             for (j = i+1 ; j <= n ; j++)
76         {
77             LL d = x[i] - x[j] ;
78             w = Cost(d) ;
79             dis[i][j] = w ;
80             dis[j][i] = w ;
81         }
82         floyed() ;
83         int u , v ;
84         while(cnt--)
85         {
86             cin>>u>>v ;
87             if (dis[u][v] != INF)
88               cout<<"The minimum cost between station "<<u<<" and station "<<v<<" is "<<dis[u][v]<<"."<<endl ;
89             else
90               cout<<"Station "<<u<<" and station "<<v<<" are not attainable."<<endl ;
91         }
92     }
93 
94     return 0 ;
95 }
View Code

 

posted @ 2015-06-21 11:35  __Meng  阅读(158)  评论(0)  编辑  收藏  举报