CF #38 Lets Go Rolling!

题意是，给出n个点对(xi, ci)，划分成若干段，每段[i..j]的代价是ci+{sum(xk-xi),i<=k<=j}。问说如何划分使得代价最小。

用DP，枚举第m+1段中的第m段，然后dp[i] =min{ dp[j] + cost(i, j) }，转移的复杂度是O(n)，状态的复杂度是O(n)，总复杂度是O(n^2)，不过用上类似单调队列优化的多重背包问题，可以把这题的复杂度优化到O(n)的。

代码

   
     

    
      
    
      #include 
    
      <
    
      iostream
    
      >
    
      
#include 
    
      <
    
      string
    
      >
    
      
#include 
    
      <
    
      stdio.h
    
      >
    
      
#include 
    
      <
    
      stdlib.h
    
      >
    
      
#include 
    
      <
    
      algorithm
    
      >
    
      
#include 
    
      <
    
      string
    
      .h
    
      >
    
      
#include 
    
      <
    
      vector
    
      >
    
      
#include 
    
      <
    
      math.h
    
      >
    
      
#include 
    
      <
    
      map
    
      >
    
      
#include 
    
      <
    
      time.h
    
      >
    
      
#include 
    
      <
    
      queue
    
      >
    
      
#include 
    
      <
    
      set
    
      >
    
      

    
      using
    
       
    
      namespace
    
       std;

typedef __int64 int64;


    
      const
    
       
    
      int
    
       MAX 
    
      =
    
       
    
      3005
    
      ;


    
      int
    
       n;
int64 cost[MAX][MAX];
int64 dp[MAX];


    
      struct
    
       NODE
{
 
    
      int
    
       x, c;
 NODE() {}
 NODE(
    
      int
    
       _x, 
    
      int
    
       _c) : x(_x), c(_c) {}
 
    
      bool
    
       
    
      operator
    
       
    
      <
    
       (
    
      const
    
       NODE 
    
      &
    
      t) 
    
      const
    
      
 {
 
    
      return
    
       x 
    
      <
    
       t.x;
 }
}node[MAX];


    
      void
    
       ready()
{
 
    
      //
    
      make the cost
    
      

    
       memset(cost, 
    
      0
    
      , 
    
      sizeof
    
      (cost));
 
    
      for
    
      (
    
      int
    
       i 
    
      =
    
       
    
      1
    
      ; i 
    
      <=
    
       n; i
    
      ++
    
      ) 
    
      for
    
      (
    
      int
    
       j 
    
      =
    
       i 
    
      +
    
       
    
      1
    
      ; j 
    
      <=
    
       n; j
    
      ++
    
      )
 {
 cost[i][j] 
    
      =
    
       cost[i][j 
    
      -
    
       
    
      1
    
      ] 
    
      +
    
       node[j].x 
    
      -
    
       node[i].x;
 
    
      //
    
      printf("%d->%d %d\n", i, j, cost[i][j]);
    
      

    
       }
}

int64 go()
{
 
    
      for
    
      (
    
      int
    
       i 
    
      =
    
       
    
      1
    
      ; i 
    
      <=
    
       n; i
    
      ++
    
      ) dp[i] 
    
      =
    
       1LL 
    
      <<
    
       
    
      60
    
      ;
 dp[
    
      0
    
      ] 
    
      =
    
       
    
      0
    
      ;
 
    
      for
    
      (
    
      int
    
       i 
    
      =
    
       
    
      1
    
      ; i 
    
      <=
    
       n; i
    
      ++
    
      )
 {
 
    
      for
    
      (
    
      int
    
       k 
    
      =
    
       
    
      0
    
      ; k 
    
      <=
    
       i 
    
      -
    
       
    
      1
    
      ; k
    
      ++
    
      )
 {
 int64 t 
    
      =
    
       dp[k] 
    
      +
    
       cost[k 
    
      +
    
       
    
      1
    
      ][i] 
    
      +
    
       node[k 
    
      +
    
       
    
      1
    
      ].c;
 
    
      if
    
      (t 
    
      <
    
       dp[i]) dp[i] 
    
      =
    
       t;
 }
 }
 
    
      return
    
       dp[n];
}


    
      int
    
       main()
{
 
    
      while
    
      (scanf(
    
      "
    
      %d
    
      "
    
      , 
    
      &
    
      n) 
    
      !=
    
       EOF)
 {
 
    
      for
    
      (
    
      int
    
       i 
    
      =
    
       
    
      1
    
      ; i 
    
      <=
    
       n; i
    
      ++
    
      ) scanf(
    
      "
    
      %d%d
    
      "
    
      , 
    
      &
    
      node[i].x, 
    
      &
    
      node[i].c);
 sort(node 
    
      +
    
       
    
      1
    
      , node 
    
      +
    
       n 
    
      +
    
       
    
      1
    
      );
 ready();
 printf(
    
      "
    
      %I64d\n
    
      "
    
      , go());
 }
}




    
      /*
    
      
3
2 3
3 4
1 2

4
1 7
3 1
5 10
6 1

    
      */