POJ 1080 Humman Gene Function

来源：互联网发布：mysql sql union 编辑：程序博客网时间：2024/04/30 18:38

大意：每两个字母都对应着一个相应的值，现在通过对短的字符串添加‘-’来求可以得到的最大值。

注意初始化dp[0][i],dp[i][0]就OK了

#include<stdio.h>#include<string.h>#include<iostream>using namespace std;int zhuanhuan(char a,char b){    if(a==b)        return 5;    else if(a=='C'&&b=='A'||a=='A'&&b=='C'||a=='T'&&b=='A'||a=='A'&&b=='T'||a=='-'&&b=='T'||a=='T'&&b=='-')        return -1;    else if(a=='A'&&b=='G'||a=='G'&&b=='A'||a=='C'&&b=='T'||a=='T'&&b=='C'||a=='G'&&b=='T'||a=='T'&&b=='G'||a=='G'&&b=='-'||a=='-'&&b=='G')        return -2;    else if(a=='A'&&b=='-'||a=='-'&&b=='A'||a=='C'&&b=='G'||a=='G'&&b=='C')        return -3;    else if(a=='C'&&b=='-'||a=='-'&&b=='C')        return -4;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        int n,m;        char s1[101],s2[101];        scanf("%d%s%d%s",&n,s1,&m,s2);        int dp[101][101];        memset(dp,0,sizeof(dp));        for(int i=1;i<=m;i++)        {            dp[0][i]=dp[0][i-1]+zhuanhuan('-',s2[i-1]);        }        for(int i=1;i<=n;i++)        {            dp[i][0]=dp[i-1][0]+zhuanhuan(s1[i-1],'-');        }        for(int i=1;i<=n;i++)        {            for(int j=1;j<=m;j++)            {                    dp[i][j]=max(max(dp[i-1][j]+zhuanhuan(s1[i-1],'-'),dp[i][j-1]+zhuanhuan('-',s2[j-1])),dp[i-1][j-1]+zhuanhuan(s1[i-1],s2[j-1]));            }        }        printf("%d\n",dp[n][m]);    }    return 0;}

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