【POJ 1080】 Human Gene Functions

【POJ 1080】 Human Gene Functions

类似于最长公共子序列的做法

dp[i][j]表示 str1[i]对应str2[j]时的最大得分

转移方程为

dp[i][j]=max(dp[i-1][j-1]+score[str1[i]][str2[j]],

max(dp[i-1][j]+score[str1[i]][‘-‘],dp[i][j-1]+score[‘-‘][str2[j]]) )

注意初始化0下标就好

代码如下:

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int dp[101][101];
char a[102],b[102];
int sc[‘Z‘+1][‘Z‘+1];

int main()
{
    int t,aa,bb,i,j;
    scanf("%d",&t);
    sc[‘A‘][‘A‘] = sc[‘C‘][‘C‘] = sc[‘G‘][‘G‘] = sc[‘T‘][‘T‘] = 5;
    sc[‘A‘][‘C‘] = sc[‘C‘][‘A‘] = sc[‘A‘][‘T‘] = sc[‘T‘][‘A‘] = sc[‘T‘][‘-‘] = -1;
    sc[‘G‘][‘A‘] = sc[‘A‘][‘G‘] = sc[‘C‘][‘T‘] = sc[‘T‘][‘C‘] = sc[‘G‘][‘T‘] = sc[‘T‘][‘G‘] = sc[‘G‘][‘-‘] = -2;
    sc[‘A‘][‘-‘] = sc[‘C‘][‘G‘] = sc[‘G‘][‘C‘] = -3;
    sc[‘C‘][‘-‘] = -4;
    while(t--)
    {
        scanf("%d %s %d %s",&aa,a+1,&bb,b+1);
        dp[0][0] = 0;
        for(i = 1; i <= bb; ++i) dp[0][i] = sc[b[i]][‘-‘] + dp[0][i-1];
        for(i = 1; i <= aa; ++i) dp[i][0] = sc[a[i]][‘-‘] + dp[i-1][0];
        for(i = 1; i <= aa; ++i)
        {
            for(j = 1; j <= bb; ++j)
            {
               dp[i][j] = max(dp[i-1][j-1] + sc[a[i]][b[j]],max(dp[i-1][j] + sc[a[i]][‘-‘],dp[i][j-1] + sc[b[j]][‘-‘]));
            }
        }
        printf("%d\n",dp[aa][bb]);
    }
    return 0;
}

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