poj 1062 昂贵的聘礼

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 15126		Accepted: 3992

      最短路问题，不过因为存在着等级的差异所以需要枚举一下。本题的思路就是对冒险者的等级进行枚举，也就是说冒险者只能和在他等级以上的人进行交易。这样枚举的好处是能够把所有的情况都考虑进去。有一点需要注意：酋长的等级不一定是最高的

代码：

 

  
    

   
     
   
      1 
   
     #include
   
     <
   
     stdio.h
   
     >
   
     

   
      2 
   
     #include
   
     <
   
     string
   
     .h
   
     >
   
     

   
      3 
   
      
   
     struct
   
      no

   
      4 
   
     {

   
      5 
   
      
   
     int
   
      p,l;

   
      6 
   
     }B[
   
     1000
   
     ];

   
      7 
   
      
   
     int
   
      map[
   
     105
   
     ][
   
     105
   
     ];

   
      8 
   
     
   
     int
   
      m,n,dist[
   
     105
   
     ],u;

   
      9 
   
     
   
     void
   
      dijkstra(
   
     int
   
      v0,
   
     int
   
      l)

   
     10 
   
     {

   
     11 
   
      
   
     int
   
      i,j,visit[
   
     105
   
     ],min;

   
     12 
   
      
   
     for
   
     (i
   
     =
   
     1
   
     ;i
   
     <=
   
     n;i
   
     ++
   
     )

   
     13 
   
      {

   
     14 
   
      
   
     if
   
     (map[v0][i]
   
     !=-
   
     1
   
      
   
     &&
   
      B[i].l
   
     -
   
     l
   
     >=
   
     0
   
      
   
     &&
   
      B[i].l
   
     -
   
     l
   
     <=
   
     m)

   
     15 
   
      dist[i]
   
     =
   
     map[v0][i];

   
     16 
   
      
   
     else
   
     

   
     17 
   
      dist[i]
   
     =
   
     0xfffffff
   
     ;

   
     18 
   
      visit[i]
   
     =
   
     0
   
     ;

   
     19 
   
      }

   
     20 
   
      visit[v0]
   
     =
   
     1
   
     ;

   
     21 
   
      
   
     for
   
     (i
   
     =
   
     1
   
     ;i
   
     <
   
     n;i
   
     ++
   
     )

   
     22 
   
      {

   
     23 
   
      min
   
     =
   
     0xfffffff
   
     ;u
   
     =
   
     i;

   
     24 
   
      
   
     for
   
     (j
   
     =
   
     1
   
     ;j
   
     <=
   
     n;j
   
     ++
   
     )

   
     25 
   
      {

   
     26 
   
      
   
     if
   
     (
   
     !
   
     visit[j] 
   
     &&
   
      dist[j]
   
     <
   
     min)

   
     27 
   
      {

   
     28 
   
      u
   
     =
   
     j;

   
     29 
   
      min
   
     =
   
     dist[j];

   
     30 
   
      }

   
     31 
   
      }

   
     32 
   
      visit[u]
   
     =
   
     1
   
     ;

   
     33 
   
      
   
     for
   
     (j
   
     =
   
     1
   
     ;j
   
     <=
   
     n;j
   
     ++
   
     )

   
     34 
   
      {

   
     35 
   
      
   
     if
   
     (
   
     !
   
     visit[j]
   
     &&
   
     (B[j].l
   
     -
   
     l)
   
     >=
   
     0
   
     &&
   
     (B[j].l
   
     -
   
     l)
   
     <=
   
     m
   
     &&
   
     map[u][j]
   
     !=-
   
     1
   
     &&
   
     dist[j]
   
     >
   
     dist[u]
   
     +
   
     map[u][j])

   
     36 
   
      dist[j]
   
     =
   
     dist[u]
   
     +
   
     map[u][j];

   
     37 
   
      }

   
     38 
   
      }

   
     39 
   
     }

   
     40 
   
     
   
     int
   
      main()

   
     41 
   
     {

   
     42 
   
      
   
     int
   
      p,l,x,j,i,b,t,min;

   
     43 
   
      scanf(
   
     "
   
     %d%d
   
     "
   
     ,
   
     &
   
     m,
   
     &
   
     n);

   
     44 
   
      memset(map,
   
     -
   
     1
   
     ,
   
     sizeof
   
     (map));

   
     45 
   
      
   
     for
   
     (i
   
     =
   
     1
   
     ;i
   
     <=
   
     n;i
   
     ++
   
     )

   
     46 
   
      {

   
     47 
   
      scanf(
   
     "
   
     %d%d%d
   
     "
   
     ,
   
     &
   
     p,
   
     &
   
     l,
   
     &
   
     x);

   
     48 
   
      
   
     for
   
     (j
   
     =
   
     1
   
     ;j
   
     <=
   
     x;j
   
     ++
   
     )

   
     49 
   
      {

   
     50 
   
      scanf(
   
     "
   
     %d%d
   
     "
   
     ,
   
     &
   
     b,
   
     &
   
     t);

   
     51 
   
      map[i][b]
   
     =
   
     t;

   
     52 
   
      }

   
     53 
   
      B[i].p
   
     =
   
     p;

   
     54 
   
      B[i].l
   
     =
   
     l;

   
     55 
   
      }

   
     56 
   
      min
   
     =
   
     B[
   
     1
   
     ].p;

   
     57 
   
      
   
     for
   
     (i
   
     =
   
     B[
   
     1
   
     ].l
   
     -
   
     m;i
   
     <=
   
     B[
   
     1
   
     ].l;i
   
     ++
   
     )

   
     58 
   
      {

   
     59 
   
      dijkstra(
   
     1
   
     ,i);

   
     60 
   
      
   
     for
   
     (j
   
     =
   
     1
   
     ;j
   
     <=
   
     n;j
   
     ++
   
     )

   
     61 
   
      {

   
     62 
   
      
   
     if
   
     (dist[j]
   
     +
   
     B[j].p
   
     <
   
     min)

   
     63 
   
      min
   
     =
   
     dist[j]
   
     +
   
     B[j].p;

   
     64 
   
      }

   
     65 
   
      }

   
     66 
   
      printf(
   
     "
   
     %d\n
   
     "
   
     ,min);

   
     67 
   
      
   
     return
   
      
   
     0
   
     ;

   
     68 
   
     }

   
     69 
   
      

   
     70 
   
     
   
     /*
   
     

   
     71 
   
     1 5

   
     72 
   
     10000 3 2

   
     73 
   
     2 5000

   
     74 
   
     3 5000

   
     75 
   
     3000 4 1

   
     76 
   
     4 400

   
     77 
   
     3000 2 1

   
     78 
   
     4 500

   
     79 
   
     1000 3 1

   
     80 
   
     5 100

   
     81 
   
     100 2 0

   
     82 
   
     

   
     83 
   
     1 5

   
     84 
   
     10000 3 2

   
     85 
   
     2 5000

   
     86 
   
     3 5000

   
     87 
   
     3000 2 1

   
     88 
   
     4 500

   
     89 
   
     3000 4 1

   
     90 
   
     4 400

   
     91 
   
     1000 3 1

   
     92 
   
     5 100

   
     93 
   
     100 2 0

   
     94 
   
     

   
     95 
   
     
   
     */
   
     

   
     96