hdu1080 dp

dp[i][j]表示配对的最大值。

dp[i-1][j]表示s1[i-1]与‘-‘配对.

dp[i][j-1]表示s2[j-1]与‘-‘配对.

dp[i-1][j-1]表示s1[i-1]与s2[j-1]配对。

dp[i][j]=max(dp[i-1][j]+mp[s1[i-1]][‘-‘],dp[i][j-1]+mp[‘-‘][s2[j-1]],dp[i-1][j-1]+mp[s1[i-1]][s2[j-1]]);

#include<stdio.h>
#include<string.h>
#define maxn 110
int dp[maxn][maxn];
int mp[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,-99999999
};

char s1[maxn],s2[maxn],c1,c2;
int cal(char c)
{
    if(c==‘A‘)return 0;
    if(c==‘C‘)return 1;
    if(c==‘G‘)return 2;
    if(c==‘T‘)return 3;
    return 4;
}
int max(int x,int y)
{
    return x>y?x:y;
}
int main()
{
    int i,j,t,l1,l2;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%s",&l1,s1);
        scanf("%d%s",&l2,s2);
        dp[0][0]=0;
        for(i=1;i<=l1;i++)
            dp[i][0]=dp[i-1][0]+mp[cal(s1[i-1])][4];
        for(i=1;i<=l2;i++)
            dp[0][i]=dp[0][i-1]+mp[4][cal(s2[i-1])];
        for(i=1;i<=l1;i++)
            for(j=1;j<=l2;j++)
            {
                dp[i][j]=dp[i-1][j-1]+mp[cal(s1[i-1])][cal(s2[j-1])];
                dp[i][j]=max(dp[i][j],dp[i-1][j]+mp[cal(s1[i-1])][4]);
                dp[i][j]=max(dp[i][j],dp[i][j-1]+mp[4][cal(s2[j-1])]);
            }
            printf("%d\n",dp[l1][l2]);
    }
}