POJ 1080 Human Gene Functions

http://poj.org/problem?id=1080

题意：

给出两段由ACGT组成的字符串，可以在字符串中插入-，将字符串进行匹配，上图为匹配值，输出最大的匹配值。

思路：

对于每一组匹配，它都有3种选择，第一是不插入-直接匹配，第二种是第一个字符串插入-再匹配，第三种是第二个字符串插入-再匹配，具体看代码。

#include<iostream>
#include<string>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;

const int maxn = 100 + 5;

int n1, n2;
char s1[maxn], s2[maxn];
int score[maxn][maxn];
int d[maxn][maxn];

void init()
{
    score['A']['A'] = 5;
    score['C']['C'] = 5;
    score['G']['G'] = 5;
    score['T']['T'] = 5;
    score['-']['-'] = -5;
    score['A']['C'] = score['C']['A'] = -1;
    score['A']['G'] = score['G']['A'] = -2;
    score['A']['T'] = score['T']['A'] = -1;
    score['A']['-'] = score['-']['A'] = -3;
    score['C']['G'] = score['G']['C'] = -3;
    score['C']['T'] = score['T']['C'] = -2;
    score['C']['-'] = score['-']['C'] = -4;
    score['G']['T'] = score['T']['G'] = -2;
    score['G']['-'] = score['-']['G'] = -2;
    score['T']['-'] = score['-']['T'] = -1;
}

int main()
{
    //freopen("D:\txt.txt", "r", stdin);
    int T;
    scanf("%d", &T);
    init();
    while (T--)
    {
        scanf("%d%s", &n1, s1 + 1);
        scanf("%d%s", &n2, s2 + 1);

        d[0][0] = 0;
        for (int i = 1; i <= n1; i++)
            d[i][0] = d[i - 1][0] + score[s1[i]]['-'];
        for (int i = 1; i <= n2; i++)
            d[0][i] = d[0][i - 1] + score['-'][s2[i]];
        for (int i = 1; i <= n1; i++)
        {
            for (int j = 1; j <= n2; j++)
            {
                d[i][j] = max(d[i-1][j] + score[s1[i]]['-'], max(d[i][j - 1] + score['-'][s2[j]],d[i-1][j-1]+score[s1[i]][s2[j]]));
            }
        }
        cout << d[n1][n2] << endl;
    }
}