HDU 1080 DP

匹配两个人相似度。

A，G，C。T。每两个都会有一个相应的值，给出两串基因。长度能够不一样，能够在基因中间加_使两串长度一样。然后有一个相应值。求最大相应值。

先做出相应的表

DP方程：

x=dp[i-1][j-1]+hash[str_a[i-1]][str_b[j-1]];

y=dp[i-1][j]+hash[str_a[i-1]]['-'];

z=dp[i][j-1]+hash[str_b[j-1]]['-'];

p[i][j]=Max(x,y,z);

#include "stdio.h"
#include "string.h"

int Max(int a,int b,int c)
{
    int x;
    x=a;
    if (b>x) x=b;
    if (c>x) x=c;
    return x;
}
int main()
{
    int Case,len_a,len_b,x,y,z,i,j;
    int dp[110][110],hash[210][210];
    char str_a[110],str_b[110];

    hash['A']['A']=hash['C']['C']=hash['T']['T']=hash['G']['G']=5;
    hash['A']['C']=hash['C']['A']=hash['A']['T']=hash['T']['A']=hash['-']['T']=hash['T']['-']=-1;
    hash['A']['G']=hash['G']['A']=hash['C']['T']=hash['T']['C']=hash['G']['T']=hash['T']['G']=hash['-']['G']=hash['G']['-']=-2;
    hash['A']['-']=hash['-']['A']=hash['C']['G']=hash['G']['C']=-3;
    hash['C']['-']=hash['-']['C']=-4;

    scanf("%d",&Case);
    while (Case--)
    {
        scanf("%d%s%d%s",&len_a,str_a,&len_b,str_b);

    //    memset(dp,0,sizeof(dp));
        dp[0][0]=0;
        for (i=1;i<=len_a;i++)
            dp[i][0]=dp[i-1][0]+hash['-'][str_a[i-1]];
        for (i=1;i<=len_b;i++)
            dp[0][i]=dp[0][i-1]+hash['-'][str_b[i-1]];

        for (i=1;i<=len_a;i++)
            for (j=1;j<=len_b;j++)
            {
                x=dp[i-1][j-1]+hash[str_a[i-1]][str_b[j-1]];
                y=dp[i-1][j]+hash[str_a[i-1]]['-'];
                z=dp[i][j-1]+hash[str_b[j-1]]['-'];
                dp[i][j]=Max(x,y,z);
            }
        printf("%d\n",dp[len_a][len_b]);
    }
    return 0;
}