poj1080

          动态规划，LCS的扩展；

            f[i][j]为a[i]与b[j]的相似值，aa[i],bb[i]是a[i],b[j]的坐标；

          动态方程是：f[i][j]=max{f[i-1][j-1]+score[aa[i-1]][bb[j-1]],f[i][j-1]+score[4][bb[j-1]],f[i-1][j]+score[aa[i-1]][4]};

         最后要注意边界：当i==0&&j==0时，f[i][j]=0;

                                          i=0,f[i][j]=f[i][j]=f[i][j-1]+score[4][bb[j-1]];

                                           j=0,f[i][j]=f[i-1][j]+score[aa[i-1]][4];

          在写代码的时候要细心,我因为弄错了一个位置,纠结了好尝试时间，也wn了好几次，呵呵。。。所以大家一定要注意啊 ！！！！

 代码如下：

               

#include<stdio.h>#include<string.h>int la,lb;char a[101],b[101];int   aa[101],bb[101];int f[101][101];int score[5][5]={{5,-1,-2,-1,-3},{-1,5,-3,-2,-4},{-2,-3,5,-2,-2},{-1,-2,-2,5,-1},{-3,-4,-2,-1,0}};void main(){   int N;   scanf("%d",&N);   while(N--)   {   int i,j,f1,f2,f3;       memset(f,0,sizeof(f));   scanf("%d%s",&la,a);   scanf("%d%s",&lb,b);   for(i=0;i<la;i++)   {       if(a[i]=='A') aa[i]=0;   if(a[i]=='C') aa[i]=1;   if(a[i]=='G') aa[i]=2;   if(a[i]=='T') aa[i]=3;   if(a[i]=='-') aa[i]=4;   }   for(i=0;i<lb;i++)   {   if(b[i]=='A') bb[i]=0;   if(b[i]=='C') bb[i]=1;   if(b[i]=='G') bb[i]=2;   if(b[i]=='T') bb[i]=3;   if(b[i]=='-') bb[i]=4;   }       for(i=0;i<=la;i++)   {   for(j=0;j<=lb;j++)   {           if(i==0 && j==0)   f[i][j]=0;   else if(i==0)   f[i][j]=f[i][j-1]+score[4][bb[j-1]];   else if(j==0)   f[i][j]=f[i-1][j]+score[aa[i-1]][4];   else   {                    f1=f[i-1][j-1]+score[aa[i-1]][bb[j-1]];f2=f[i][j-1]+score[4][bb[j-1]];f3=f[i-1][j]+score[aa[i-1]][4];f[i][j]=(f1>f2?f1:f2)>f3?(f1>f2?f1:f2):f3;   }   }   }  printf("%d\n",f[la][lb]);   }}

        希望对大家有一定的帮助！！！！！