C++ OJ HDU 1159 Common Subsequence

C++ OJ 

HDU 1159 Common Subsequence

CLAY 撰写  XCOJ  动态规划 暂无评论

Now I think you have got an AC in Ignatius.L’s “Max Sum” problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem.

Given a consecutive number sequence S1, S2, S3, S4 … Sx, … Sn (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ Sx ≤ 32767). We define a function sum(i, j) = Si + … + Sj (1 ≤ i ≤ j ≤ n).

Now given an integer m (m > 0), your task is to find m pairs of i and j which make sum(i1, j1) + sum(i2, j2) + sum(i3, j3) + … + sum(im, jm) maximal (ix ≤ iy ≤ jx or ix ≤ jy ≤ jxis not allowed).

But I`m lazy, I don’t want to write a special-judge module, so you don’t have to output m pairs of i and j, just output the maximal summation of sum(ix, jx)(1 ≤ x ≤ m) instead. ^_^

输入

Each test case will begin with two integers m and n, followed by n integers S1, S2, S3 … Sn.
Process to the end of file.

输出

Output the maximal summation described above in one line.

样例输入

1 2	1 3 1 2 3 2 6 - 1 4 - 2 3 - 2 3

样例输出

1 2

6 8

	a	b	c	f	b	c
a	1	1	1	1	1	1
b	1	2	2	2	2	2
f	1	2	2	3	3	3
c	1	2	3	3	3	4
a	1	2	3	3	3	4
b	1	2	3	3	4	4

大概就是画这样一个表格，动态规划的方式去解题

F[i][j]=F[i-1][j-1]+1；（a[i]==b[j]）

F[i][j]=max(F[i-1][j],F[i][j-1])(a[i]!=b[j]);

n由于F(i,j)只和F(i-1,j-1), F(i-1,j)和F(i,j-1)有关, 而在计算F(i,j)时, 只要选择一个合适的顺序, 就可以保证这三项都已经计算出来了, 这样就可以计算出F(i,j). 这样一直推到f(len(a),len(b))就得到所要求的解了.

其实自己还有一种想法不过也和这个差不多

	a	b	c	f	b	c
a	1	0	0	0	0	0
b	0	1	0	0	1	0
f	0	0	0	1	0	0
c	0	0	1	0	0	1
a	1	0	0	0	0	0
b	0	1	0	0	1	0

算法是只能往下，然后找到一个出表格的最长的路，其实本质上差不多

下面贴代码

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#include<stdio.h> #include<string.h> #define MAX 1001 int F [ MAX ] [ MAX ] ; int main ( ) { char X [ MAX ] ; char Y [ MAX ] ;      int lenx = 0 , leny = 0 , i , j ;      while ( ~ scanf ( "%s %s" , X , Y ) )      {          lenx = strlen ( X ) ;          leny = strlen ( Y ) ;          for ( i = 0 ; i <= lenx ; i ++ )          {            F [ i ] [ 0 ] = 0 ;          }          for ( i = 0 ; i <= leny ; i ++ )          {              F [ 0 ] [ i ] = 0 ;          }          for ( i = 1 ; i <= lenx ; i ++ )          {              for ( j = 1 ; j <= leny ; j ++ )              {                  if ( X [ i - 1 ] == Y [ j - 1 ] )                  {                      F [ i ] [ j ] = F [ i - 1 ] [ j - 1 ] + 1 ;                  }                  else                  {                      F [ i ] [ j ] = F [ i - 1 ] [ j ] > F [ i ] [ j - 1 ] ? F [ i - 1 ] [ j ] : F [ i ] [ j - 1 ] ;                  }              }          }          printf ( "%d\n" , F [ lenx ] [ leny ] ) ;      }      return 0 ; }