poj 1080 Human Gene Functions

针对三种情况进行最大值的求解，规模为100，只有使用动态规划。

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <vector>#include <cmath>#include <algorithm>#include <set>#include <map>#include <queue>#include <ctime>#define pb push_back#define ll long long#define mp make_pair#define f first#define s second#define pii pair < int, int >#define ull unsigned long long#define pll pair < ll, ll >#define forit(s) for(__typeof(s.begin()) it = s.begin(); it != s.end(); it ++)#define all(s) s.begin(), s.end()const int inf = (1ll << 31) - 1;const int maxn = (int) 1e5 + 10;#define max(a, b) ((a) > (b) ? (a) : (b))#define min(a, b) ((a) < (b) ? (a) : (b))using namespace std;int matrix[5][5] = {{-19931117, -3, -4, -2, -1},                    {-3, 5, -1, -2, -1},                    {-4, -1, 5, -3, -2},                    {-2, -2, -3, 5, -2},                    {-1, -1, -2, -2, 5}};int dp[110][110];char s1[110], s2[110];int T = 0, len1 = 0, len2 = 0;int change(char a){    if (a == 'A')    {        return 1;    }else if (a == 'C')    {        return 2;    }else if (a == 'G')    {        return 3;    }else if (a == 'T')    {        return 4;    }else     {        return 0;    }}int comp(char a, char b){    int r = 0;    int c = 0;    r = change(a);    c = change(b);    return matrix[r][c];    } int main(){    //freopen("data_1080.txt", "r", stdin);    while(scanf("%d", &T) != EOF)    {        for (int t = 0; t < T; t++)        {            memset(dp, 0, sizeof(dp));            scanf("%d%s", &len1, s1);            scanf("%d%s", &len2, s2);            dp[0][0] = 0;            char *s11, *s22;            s11= s1 - 1;            s22 = s2 - 1;            for (int i = 1; i <= len1; i++)            {                dp[i][0] = dp[i - 1][0] + comp(s11[i], 0);            }            for (int i = 1; i <= len2; i++)            {                dp[0][i] = dp[0][i - 1] + comp(s22[i], 0);            }            for (int i = 1; i <= len1; i++)            {                for (int j = 1; j <= len2; j++)                {                    dp[i][j] = dp[i][j - 1] + comp(s22[j], 0);                    dp[i][j] = max(dp[i][j], dp[i-1][j] + comp(s11[i], 0));                    dp[i][j] = max(dp[i][j], dp[i-1][j-1] + comp(s11[i], s22[j]));                }            }            printf("%d\n", dp[len1][len2]);        }    }    return 0;}