HDU 1080 POJ 1080 Human Gene Functions——動態規劃

http://poj.org/problem?id=1080
http://acm.hdu.edu.cn/showproblem.php?pid=1080

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

#define MAX(x,y,z) ((x)>(y)?((x)>(z)?(x):(z)):((y)>(z)?(y):(z)))

using namespace std;

char str1[105];
char str2[105];
int list[6][6] = {
//          { \0,  A,  C,  G,  T,  -}
    /*\0*/  {  0,  0,  0,  0,  0,  0},
    /*A*/   {  0,  5, -1, -2, -1, -3},
    /*C*/   {  0, -1,  5, -3, -2, -4},
    /*G*/   {  0, -2, -3,  5, -2, -2},
    /*T*/   {  0, -1, -2, -2,  5, -1},
    /*-*/   {  0, -3, -4, -2, -1,  0},
};
map<char, int>m;

int dp[105][105];//dp[x][y]表示str1中1...x-1個字符和str2中第1...y-1之間匹配的最優解

int main(int argc, char const *argv[])
{
    int T, len1, len2;
    scanf("%d", &T);

    m['\0'] = 0;
    m['A'] = 1;
    m['C'] = 2;
    m['G'] = 3;
    m['T'] = 4;
    m['-'] = 5;

    while (T--) {
        memset(dp, 0, sizeof(dp));
        scanf("%d%s", &len1, str1 + 1);
        scanf("%d%s", &len2, str2 + 1);

        // printf("\t\t");
        // for (int y = 0; y < len2; y++) {
        //  printf("%c\t", s2[y]);
        // }
        // printf("\n\t");

        // for (int y = 0; y <= len2; y++) {
        //  printf("%d\t", dp[0][y]);

        // }
        // printf("\n");

        for (int x = 1; x <= len1 + 1; x++) {
            dp[x][0] = -1e9;
        }

        for (int y = 1; y <= len2 + 1; y++) {
            dp[0][y] = -1e9;
        }

        for (int x = 1; x <= len1 + 1; x++) {

            for (int y = 1; y <= len2 + 1; y++) {
                dp[x][y] = MAX(
                               dp[x - 1][y - 1] + list[m[str1[x - 1]]][m[str2[y - 1]]],
                               dp[x - 1][y] + list[m[str1[x - 1]]][m['-']],
                               dp[x][y - 1] + list[m['-']][m[str2[y - 1]]]
                           );

//取消注釋打印流程

/*              if (dp[x - 1][y - 1] + list[m[str1[x - 1]]][m[str2[y - 1]]]
                        >= dp[x - 1][y] + list[m[str1[x - 1]]][m['-']]) {
                    if (dp[x - 1][y - 1] + list[m[str1[x - 1]]][m[str2[y - 1]]]
                            >=
                            dp[x][y - 1] + list[m['-']][m[str2[y - 1]]]) {
                        printf("↖%d\t", dp[x][y]);
                    } else {
                        printf("←%d\t", dp[x][y]);
                    }
                } else {
                    if (dp[x - 1][y] + list[m[str1[x - 1]]][m['-']]
                            >=
                            dp[x][y - 1] + list[m['-']][m[str2[y - 1]]]) {
                        printf("↑%d\t", dp[x][y]);
                    } else {
                        printf("←%d\t", dp[x][y]);

                    }
                }
*/

            }

/*
            printf("\n");
*/

        }
        printf("%d\n", dp[len1 + 1][len2 + 1]);

        /**
        *
        *   in:
        *   
        *   2
        *   7 AGTGATG
        *   5 GTTAG
        *   7 AGCTATT
        *   9 AGCTTTAAA
        *   
        *   out:
        *   
        *   ↖0      ←-2     ←-3     ←-4     ←-7     ←-9
        *   ↑-3     ↖-2     ↖-3     ↖-4     ↖1      ←-1
        *   ↑-5     ↖2      ←1      ←0      ↑-1     ↖6
        *   ↑-6     ↑1      ↖7      ↖6      ←3      ↑5
        *   ↑-8     ↖-1     ↑5      ↖5      ↖4      ↖8
        *   ↑-11    ↑-4     ↑2      ↖4      ↖10     ←8
        *   ↑-12    ↑-5     ↖1      ↖7      ↑9      ↖8
        *   ↑-14    ↖-7     ↑-1     ↑5      ↑7      ↖14
        *   
        *   14
        *   
        *   ↖0      ←-3     ←-5     ←-9     ←-10    ←-11    ←-12    ←-15    ←-18    ←-21
        *   ↑-3     ↖5      ←3      ←-1     ←-2     ←-3     ←-4     ↖-7     ↖-10    ↖-13
        *   ↑-5     ↑3      ↖10     ←6      ←5      ←4      ←3      ←0      ←-3     ←-6
        *   ↑-9     ↑-1     ↑6      ↖15     ←14     ←13     ←12     ←9      ←6      ←3
        *   ↑-10    ↑-2     ↑5      ↑14     ↖20     ↖19     ↖18     ←15     ←12     ←9
        *   ↑-13    ↖-5     ↑2      ↑11     ↑17     ↖19     ↖18     ↖23     ↖20     ↖17
        *   ↑-14    ↑-6     ↑1      ↑10     ↖16     ↖22     ↖24     ↑22     ↖22     ↖19
        *   ↑-15    ↑-7     ↑0      ↑9      ↖15     ↖21     ↖27     ←24     ↖21     ↖21
        *   
        *   
        *   21
        *   
        *   
        *   
        */

    }
    return 0;
}

後來發現是自己想得太多了，這個題的“狀態”不一定非要理解成原先那樣，
其實完全可以也像最大公共子列那樣的，
dp[x][y]表示str1的前x個字符和str2的前y個字符之間的匹配結果的最優解
這樣也便於理解，便於思考

這樣的話dp[x][y]就是最終答案，不過要注意邊界的預處理
代碼如下

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

#define MAX(x,y,z) ((x)>(y)?((x)>(z)?(x):(z)):((y)>(z)?(y):(z)))

using namespace std;

char str1[105];
char str2[105];
int list[6][6] = {
//          {  A,  C,  G,  T,  -}
    /*A*/   {  5, -1, -2, -1, -3},
    /*C*/   { -1,  5, -3, -2, -4},
    /*G*/   { -2, -3,  5, -2, -2},
    /*T*/   { -1, -2, -2,  5, -1},
    /*-*/   { -3, -4, -2, -1,  0},
};

map<char, int>m;

int dp[105][105];//dp[x][y]表示str1中1...x個字符和str2中第1...y這兩個子串之間的所有匹配方式的最大利益

int main(int argc, char const *argv[])
{
    int T, len1, len2;
    scanf("%d", &T);

    m['A'] = 0;
    m['C'] = 1;
    m['G'] = 2;
    m['T'] = 3;
    m['-'] = 4;

    while (T--) {
        memset(dp, 0, sizeof(dp));
        scanf("%d%s", &len1, str1 + 1);
        scanf("%d%s", &len2, str2 + 1);

        // printf("\t\t");
        // for (int y = 0; y < len2; y++) {
        //  printf("%c\t", s2[y]);
        // }
        // printf("\n\t");

        // for (int y = 0; y <= len2; y++) {
        //  printf("%d\t", dp[0][y]);

        // }
        // printf("\n");

        for (int x = 1; x <= len1; x++) {
            dp[x][0] = dp[x - 1][0] + list[m[str1[x]]][m['-']];
        }

        // printf("↖0\t");
        for (int y = 1; y <= len2; y++) {
            dp[0][y] = dp[0][y - 1] + list[m[str2[y]]][m['-']];
            // printf("←%d\t", dp[0][y]);
        }
        // printf("\n");

        for (int x = 1; x <= len1; x++) {

            // printf("↑%d\t", dp[x][0]);

            for (int y = 1; y <= len2; y++) {
                dp[x][y] = MAX(
                               dp[x - 1][y - 1] + list[m[str1[x]]][m[str2[y]]],
                               dp[x - 1][y] + list[m[str1[x]]][m['-']],
                               dp[x][y - 1] + list[m['-']][m[str2[y]]]
                           );

//取消所有額外註釋可打印流程
                /*
                                if (dp[x - 1][y - 1] + list[m[str1[x]]][m[str2[y]]]
                                        >= dp[x - 1][y] + list[m[str1[x]]][m['-']]) {
                                    if (dp[x - 1][y - 1] + list[m[str1[x]]][m[str2[y]]]
                                            >=
                                            dp[x][y - 1] + list[m['-']][m[str2[y]]]) {
                                        printf("↖%d\t", dp[x][y]);
                                    } else {
                                        printf("←%d\t", dp[x][y]);
                                    }
                                } else {
                                    if (dp[x - 1][y] + list[m[str1[x]]][m['-']]
                                            >=
                                            dp[x][y - 1] + list[m['-']][m[str2[y]]]) {
                                        printf("↑%d\t", dp[x][y]);
                                    } else {
                                        printf("←%d\t", dp[x][y]);

                                    }
                                }
                */

            }

            /*
                        printf("\n");
            */

        }
        printf("%d\n", dp[len1][len2]);

        /**
        *
        *   in:
        *
        *   2
        *   7 AGTGATG
        *   5 GTTAG
        *   7 AGCTATT
        *   9 AGCTTTAAA
        *
        *   out:
        *
        *   ↖0      ←-2     ←-3     ←-4     ←-7     ←-9
        *   ↑-3     ↖-2     ↖-3     ↖-4     ↖1      ←-1
        *   ↑-5     ↖2      ←1      ←0      ↑-1     ↖6
        *   ↑-6     ↑1      ↖7      ↖6      ←3      ↑5
        *   ↑-8     ↖-1     ↑5      ↖5      ↖4      ↖8
        *   ↑-11    ↑-4     ↑2      ↖4      ↖10     ←8
        *   ↑-12    ↑-5     ↖1      ↖7      ↑9      ↖8
        *   ↑-14    ↖-7     ↑-1     ↑5      ↑7      ↖14
        *
        *   14
        *
        *   ↖0      ←-3     ←-5     ←-9     ←-10    ←-11    ←-12    ←-15    ←-18    ←-21
        *   ↑-3     ↖5      ←3      ←-1     ←-2     ←-3     ←-4     ↖-7     ↖-10    ↖-13
        *   ↑-5     ↑3      ↖10     ←6      ←5      ←4      ←3      ←0      ←-3     ←-6
        *   ↑-9     ↑-1     ↑6      ↖15     ←14     ←13     ←12     ←9      ←6      ←3
        *   ↑-10    ↑-2     ↑5      ↑14     ↖20     ↖19     ↖18     ←15     ←12     ←9
        *   ↑-13    ↖-5     ↑2      ↑11     ↑17     ↖19     ↖18     ↖23     ↖20     ↖17
        *   ↑-14    ↑-6     ↑1      ↑10     ↖16     ↖22     ↖24     ↑22     ↖22     ↖19
        *   ↑-15    ↑-7     ↑0      ↑9      ↖15     ↖21     ↖27     ←24     ↖21     ↖21
        *
        *
        *   21
        *
        *
        *
        */

    }
    return 0;
}