HDU 1102 Constructing Roads

基础的最小生成树 有的路已经建好了，就把矩阵置0 就可以了

  
    

   
     //
   
     prim MST
   
     

   
     #include 
   
     <
   
     stdio.h
   
     >
   
     
#include 
   
     <
   
     stdlib.h
   
     >
   
     
#include 
   
     <
   
     string
   
     .h
   
     >
   
     

   
     int
   
      G[
   
     101
   
     ][
   
     101
   
     ]; 

   
     bool
   
      used[
   
     101
   
     ]; 

   
     int
   
      lowcost[
   
     101
   
     ];

   
     int
   
      closest[
   
     101
   
     ];

   
     int
   
      cost;


   
     void
   
      prim(
   
     int
   
      n){
 cost 
   
     =
   
      
   
     0
   
     ;
 used[
   
     1
   
     ] 
   
     =
   
      
   
     true
   
     ;
 
   
     int
   
      minj;
 
   
     for
   
     (
   
     int
   
      i 
   
     =
   
      
   
     2
   
     ; i 
   
     <=
   
      n; i
   
     ++
   
     ){
 lowcost[i] 
   
     =
   
      G[i][
   
     1
   
     ];
 closest[i] 
   
     =
   
      
   
     1
   
     ;
 used[i] 
   
     =
   
      
   
     false
   
     ;
 }
 
   
     for
   
     (
   
     int
   
      i 
   
     =
   
      
   
     1
   
     ; i 
   
     <
   
      n; i
   
     ++
   
     ){ 
   
     //
   
     n-1次
   
     

   
      
   
     int
   
      min 
   
     =
   
      
   
     32767
   
     ;
 
   
     for
   
     (
   
     int
   
      k 
   
     =
   
      
   
     1
   
     ; k 
   
     <=
   
      n; k
   
     ++
   
     ){
 
   
     if
   
     ( 
   
     !
   
     used[k] 
   
     &&
   
      lowcost[k] 
   
     <
   
      min ){
 min 
   
     =
   
      lowcost[k];
 minj 
   
     =
   
      k; 
 }
 }
 used[minj] 
   
     =
   
      
   
     true
   
     ;
 cost 
   
     +=
   
      min;
 
   
     for
   
     (
   
     int
   
      j 
   
     =
   
      
   
     1
   
     ; j 
   
     <=
   
      n;j
   
     ++
   
     ){
 
   
     if
   
     ( 
   
     !
   
     used[j] 
   
     &&
   
      G[minj][j] 
   
     <
   
      lowcost[j])
 lowcost[j] 
   
     =
   
      G[minj][j];
 }
 }
}


   
     int
   
      main(){
 
   
     int
   
      N,Q,x,y;
 
   
     while
   
     (scanf(
   
     "
   
     %d
   
     "
   
     ,
   
     &
   
     N) 
   
     ==
   
      
   
     1
   
     ){
 
   
     for
   
     (
   
     int
   
      i 
   
     =
   
      
   
     1
   
     ; i 
   
     <=
   
      N;i
   
     ++
   
     )
 
   
     for
   
     (
   
     int
   
      j 
   
     =
   
      
   
     1
   
     ; j 
   
     <=
   
      N; j
   
     ++
   
     )
 scanf(
   
     "
   
     %d
   
     "
   
     , 
   
     &
   
     G[i][j]);
 scanf(
   
     "
   
     %d
   
     "
   
     ,
   
     &
   
     Q);
 
   
     //
   
     有连线的 将G[x][y] 置0
   
     

   
      
   
     while
   
     (Q
   
     --
   
     ){
 scanf(
   
     "
   
     %d %d
   
     "
   
     ,
   
     &
   
     x,
   
     &
   
     y);
 G[x][y] 
   
     =
   
      
   
     0
   
     ;
 G[y][x] 
   
     =
   
      
   
     0
   
     ;
 }
 prim(N);
 printf(
   
     "
   
     %d\n
   
     "
   
     ,cost);
 }
 
   
     return
   
      
   
     0
   
     ;
}