HDU 1080 Human Gene Functions

最长公共子序列的变形题，算法导论公共子序列后的练习题有题是类似的

注意在公共子序列中要判断a[i]和b[j]相等，而此题是考虑相似，不要求相等

#include <iostream>
#include <vector>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <string>
#include <map>

#define MAX 100 + 5
using namespace std;

//dp[i][j]表示字符串a以a[i]结尾，字符串b以b[j]结尾的最大相似度
int dp[MAX][MAX];

int score[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 main(){
    map<char,int> table;
    table['A'] = 0;table['C'] = 1;table['G'] = 2;
    table['T'] = 3;table['-'] = 4;
    int T;
    cin >> T;
    while(T--){
        int na,nb;
        string stra,strb;
        cin >> na>>stra >> nb>>strb;
        memset(dp,0,sizeof(dp));
        //注意边界条件
        for(int i = 1; i <= na; i ++ ) dp[i][0] = dp[i-1][0] + score[ table[ stra[i-1] ] ][ table[ '-' ] ];
        for(int i = 1; i <= nb; i ++ ) dp[0][i] = dp[0][i-1] + score[ table['-'] ][ table[ strb[i-1] ] ];

        for(int i = 1 ; i <= na; i ++ ){
            for(int  j = 1; j <= nb; j ++){
                dp[i][j]= max(dp[i- 1  ][j -1 ] + score[ table[ stra[i-1] ] ][ table[ strb[j-1] ]],
                              max(dp[i- 1  ][j] + score[ table[ stra[i-1] ]][ table[ '-' ]],  dp[i][j-1] + score[ table[ '-' ]][ table[ strb[j-1] ]]) );
            }
        }
        cout<<dp[na][nb]<<endl;

    }


    return 0;
}