ZOJ1027

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=27

类似于最长公共子序列。

用f[i][j]表示第一个基因序列s1的前i个字符和第二个基因序列s2的前j个字符对齐后的最大相似度，函数dis用来计算两个字符之间的相似度，分三种情况：

1.在s1的第i个字符后面插入空格与s2的第j个字符匹配，f[i][j] = f[i][j-1]+dis(s2[j],'-')；

2.在s2的第j个字符后面插入空格与s1的第j个字符匹配，f[i][j] = f[i-1][j]+dis(s1[i],'-')；

3.让s1的第i个字符与s2的第j个字符匹配，f[i][j] = f[i-1][j-1]+dis(s1[i],s2[j])。

注意考虑初始化：

for (int i=1; i<=n; i++)

           f[i][0] = f[i-1][0] + dis(s1[i],'-');
for (int j=1; j<=m; j++)
            f[0][j] = f[0][j-1] + dis(s2[j],'-');

#include<iostream>
#include<cstdio>
#include<memory.h>
using namespace std;

int f[110][110];
int d[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,-100}};

int dis(char c1, char c2)
{
    int a, b;
    switch (c1)
    {
        case 'A':a = 0; break;
        case 'C':a = 1; break;
        case 'G':a = 2; break;
        case 'T':a = 3; break;
        case '-':a = 4; break;
    }
    switch (c2)
    {
        case 'A':b = 0; break;
        case 'C':b = 1; break;
        case 'G':b = 2; break;
        case 'T':b = 3; break;
        case '-':b = 4; break;
    }
    return d[a][b];
}

int main()
{
    int T;

    scanf("%d",&T);
    while (T--)
    {
        int n,m;
        char s1[110],s2[110];
        scanf("%d%s",&n,&s1[1]);
        scanf("%d%s",&m,&s2[1]);
        memset(f,0,sizeof(0));
        for (int i=1; i<=n; i++)
            f[i][0] = f[i-1][0] + dis(s1[i],'-');
        for (int j=1; j<=m; j++)
            f[0][j] = f[0][j-1] + dis(s2[j],'-');
        for (int i=1; i<=n; i++)
            for (int j=1; j<=m; j++)
        {
            f[i][j] = max(max(f[i][j-1]+dis(s2[j],'-'),
                              f[i-1][j]+dis(s1[i],'-')),
                          f[i-1][j-1]+dis(s1[i],s2[j]));
        }
        cout<<f[n][m]<<endl;
    }

    return 0;
}