hdu2048

带路径的数塔

   
     

    
      #include
    
      "
    
      iostream
    
      "
    
      

    
      using
    
       
    
      namespace
    
       std;

    
      int
    
       main()
{
 
    
      int
    
       a[
    
      100
    
      ][
    
      100
    
      ],b[
    
      100
    
      ];
 
    
      int
    
       i,j;
 
    
      int
    
       n;
 
    
      while
    
      (cin
    
      >>
    
      n)
 {
 
    
      for
    
      (i
    
      =
    
      0
    
      ;i
    
      <
    
      n;i
    
      ++
    
      )
 
    
      for
    
      (j
    
      =
    
      0
    
      ;j
    
      <=
    
      i;j
    
      ++
    
      )
 cin
    
      >>
    
      a[i][j];
 
    
      int
    
       t
    
      =
    
      0
    
      ;
 
    
      int
    
       y
    
      =
    
      a[
    
      0
    
      ][
    
      0
    
      ];
 
    
      for
    
      (i
    
      =
    
      n
    
      -
    
      2
    
      ;i
    
      >=
    
      0
    
      ;i
    
      --
    
      )
 {
 
    
      int
    
       max
    
      =
    
      0
    
      ,k
    
      =
    
      0
    
      ;
 
    
      for
    
      (j
    
      =
    
      0
    
      ;j
    
      <=
    
      i;j
    
      ++
    
      )
 {
 
    
      if
    
      (a[i
    
      +
    
      1
    
      ][j]
    
      >
    
      a[i
    
      +
    
      1
    
      ][j
    
      +
    
      1
    
      ])//将每次出现的最大路径保存即可
 {
 a[i][j]
    
      +=
    
      a[i
    
      +
    
      1
    
      ][j]; 
 
    
      if
    
      (max
    
      <
    
      a[i][j]) { max
    
      =
    
      a[i][j]; k
    
      =
    
      a[i
    
      +
    
      1
    
      ][j]; b[t
    
      ++
    
      ]
    
      =
    
      k;}
 }
 
    
      else
    
       {
 a[i][j]
    
      +=
    
      a[i
    
      +
    
      1
    
      ][j
    
      +
    
      1
    
      ]; 
 
    
      if
    
      (max
    
      <
    
      a[i][j]) {max
    
      =
    
      a[i][j]; k
    
      =
    
      a[i
    
      +
    
      1
    
      ][j
    
      +
    
      1
    
      ]; b[t
    
      ++
    
      ]
    
      =
    
      k;}
 
 }
 }
 }
 cout
    
      <<
    
      a[
    
      0
    
      ][
    
      0
    
      ]
    
      <<
    
      endl;
 cout
    
      <<
    
      y
    
      <<
    
      "
    
       
    
      "
    
      ;
 
    
      if
    
      (n
    
      %
    
      2
    
      ==
    
      0
    
      )
 {
 
    
      for
    
      (i
    
      =
    
      t
    
      -
    
      1
    
      ;i
    
      >=
    
      0
    
      ;i
    
      --
    
      ) //求路径之差
 {
 cout
    
      <<
    
      b[i]
    
      -
    
      b[i
    
      -
    
      1
    
      ];
 
    
      if
    
      (i
    
      !=
    
      0
    
      ) cout
    
      <<
    
      "
    
       
    
      "
    
      ;
 }
 }
 
    
      else
    
      
 {
 
    
      for
    
      (i
    
      =
    
      t
    
      -
    
      1
    
      ;i
    
      >=
    
      2
    
      ;i
    
      --
    
      )
 {
 cout
    
      <<
    
      b[i]
    
      -
    
      b[i
    
      -
    
      1
    
      ];
 
    
      if
    
      (i
    
      !=
    
      2
    
      ) cout
    
      <<
    
      "
    
       
    
      "
    
      ;
 }
 cout
    
      <<
    
      "
    
       
    
      "
    
      <<
    
      b[
    
      0
    
      ];
 }
 
    
      //
    
      cout<<endl;
    
      

    
      
 }
 
    
      return
    
       
    
      0
    
      ;
}
   
     

数塔（典型）

   
     

    
       1 
    
      #include
    
      "
    
      iostream
    
      "
    
      

    
       2 
    
      
    
      #define
    
       M 100
    
      

    
       3 
    
      
    
      using
    
       
    
      namespace
    
       std;

    
       4 
    
      
    
      int
    
       Max(
    
      int
    
       a,
    
      int
    
       b)

    
       5 
    
      {

    
       6 
    
       
    
      return
    
       a
    
      >
    
      b
    
      ?
    
      a:b;

    
       7 
    
      }

    
       8 
    
      
    
      int
    
       main()

    
       9 
    
      {

    
      10 
    
       
    
      int
    
       m,n;

    
      11 
    
       
    
      int
    
       i,j;

    
      12 
    
       
    
      int
    
       dp[M][M];

    
      13 
    
       cin
    
      >>
    
      m;

    
      14 
    
       
    
      while
    
      (m
    
      --
    
      )

    
      15 
    
       {

    
      16 
    
       cin
    
      >>
    
      n;

    
      17 
    
       

    
      18 
    
       
    
      for
    
      (i
    
      =
    
      0
    
      ;i
    
      <
    
      n;i
    
      ++
    
      )

    
      19 
    
       
    
      for
    
      (j
    
      =
    
      0
    
      ;j
    
      <=
    
      i;j
    
      ++
    
      )

    
      20 
    
       cin
    
      >>
    
      dp[i][j];

    
      21 
    
      

    
      22 
    
       
    
      for
    
      (i
    
      =
    
      n
    
      -
    
      2
    
      ;i
    
      >=
    
      0
    
      ;i
    
      --
    
      )

    
      23 
    
       {

    
      24 
    
       
    
      for
    
      (j
    
      =
    
      0
    
      ;j
    
      <
    
      n;j
    
      ++
    
      )

    
      25 
    
       dp[i][j]
    
      +=
    
      Max(dp[i
    
      +
    
      1
    
      ][j],dp[i
    
      +
    
      1
    
      ][j
    
      +
    
      1
    
      ]);

    
      26 
    
       }

    
      27 
    
       cout
    
      <<
    
      dp[
    
      0
    
      ][
    
      0
    
      ]
    
      <<
    
      endl;

    
      28 
    
       }

    
      29 
    
       
    
      return
    
       
    
      0
    
      ;

    
      30 
    
      }