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

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1 3 1 2 3

2 6 -1 4 -2 3 -2 3

样例输出

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6

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 abcfbca111111b122222f122333c123334a123334b123344

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

 abcfbca100000b010010f000100c001001a100000b010010

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

intF[MAX][MAX];

intmain()

{

charX[MAX];

charY[MAX];

    intlenx=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]);

    }

    return0;

}