POJ 1080 Human Gene Functions

来源：互联网发布：win10版qq知乎编辑：程序博客网时间：2024/05/02 01:46

这题方程很好想，只是我觉得不太好证明最优子结构。。。智商捉急。。。

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

Discuss里面有很好很强大的证明。。。

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int mt[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}};int conv(char tc){    switch (tc)    {        case 'A':            return 0;        case 'C':            return 1;        case 'G':            return 2;        case 'T':            return 3;    }}int main(){    int i,j,prob,l1,l2,s1[110],s2[110],dp[110][100],t;    char str[110];    scanf("%d",&prob);    while (prob--)    {        scanf("%d",&l1);        scanf("%s",str);        for (i=0; i<l1; i++)        {            s1[i+1]=conv(str[i]);        }        scanf("%d",&l2);        scanf("%s",str);        for (i=0; i<l2; i++)        {            s2[i+1]=conv(str[i]);        }        dp[0][0]=0;        for (i=1; i<=l1; i++)        {            dp[i][0]=dp[i-1][0]+mt[s1[i]][4];        }        for (i=1; i<=l2; i++)        {            dp[0][i]=dp[0][i-1]+mt[s2[i]][4];        }        for (i=1; i<=l1; i++)        {            for (j=1; j<=l2; j++)            {                    t=dp[i-1][j]+mt[s1[i]][4];                    dp[i][j]=t;                    t=dp[i][j-1]+mt[s2[j]][4];                    dp[i][j]=dp[i][j]>t?dp[i][j]:t;                    t=dp[i-1][j-1]+mt[s1[i]][s2[j]];                    dp[i][j]=dp[i][j]>t?dp[i][j]:t;            }        }        printf("%d\n",dp[l1][l2]);    }}