SOJ-2708（简单dp，不会怎么记录路径）

参考了别人记录结果路径的方法，其余为自己所想

double a[30];
double d[30][30];
double ans[10005][30];
int path[10005][30];
char s[10000];
void process(char c)
{
    int i;
    switch (c) {
        case '2':
            for (i = 0; i < 3; ++i)
                ans[0][i] = a[i];
            break;
        case '3':
            for (i = 3; i < 6; ++i)
                ans[0][i] = a[i];
            break;
        case '4':
            for (i = 6; i < 9; ++i)
                ans[0][i] = a[i];
            break;
        case '5':
            for (i = 9; i < 12; ++i)
                ans[0][i] = a[i];
            break;
        case '6':
            for (i = 12; i < 15; ++i)
                ans[0][i] = a[i];
            break;
        case '7':
            for (i = 15; i < 19; ++i)
                ans[0][i] = a[i];
            break;
        case '8':
            for (i = 19; i < 22; ++i)
                ans[0][i] = a[i];
            break;
        case '9':
            for (i = 22; i < 26; ++i)
                ans[0][i] = a[i];
            break;
    }
}
bool check(int a, char b)
{
    if (a >= 0 && a <= 2 && b == '2') return true;
    if (a >= 3 && a <= 5 && b == '3') return true;
    if (a >= 6 && a <= 8 && b == '4') return true;
    if (a >= 9 && a <= 11 && b == '5') return true;
    if (a >= 12 && a <= 14 && b == '6') return true;
    if (a >= 15 && a <= 18 && b == '7') return true;
    if (a >= 19 && a <= 21 && b == '8') return true;
    if (a >= 22 && a <= 25 && b == '9') return true;
    return false;
}
void easy(int a, int b, int in)
{
    int i, j;
    for (i = a; i < b; ++i) {
        double max = -1;
        int index = 0;
        for (j = 0; j < 26; ++j) {
            if ((d[j][i]) > 0 && (ans[in - 1][j]) > 0 && check(j, s[in - 1])) {
                if (ans[in - 1][j] * d[j][i] > max) {
                    max = ans[in - 1][j] * d[j][i];
                    path[in][i] = j;
                }
            }
        }
        if (max == -1) max = 0;
        ans[in][i] = max;
    }
}
void process1(char c, int index)
{
    int i, j;
    switch (c) {
        case '2':
            easy(0, 3, index);
            break;
        case '3':
            easy(3, 6, index);
            break;
        case '4':
            easy(6, 9, index);
            break;
        case '5':
            easy(9, 12, index);
            break;
        case '6':
            easy(12, 15, index);
            break;
        case '7':
            easy(15, 19, index);
            break;
        case '8':
            easy(19, 22, index);
            break;
        case '9':
            easy(22, 26, index);
            break;
    }
}
int main()
{
    int i, j;
    for (i = 0; i < 26; ++i)
        scanf("%lf", &a[i]);
    for (i = 0; i < 26; ++i) {
        for (j = 0; j < 26; ++j) {
            scanf("%lf", &d[i][j]);
        }
    }
    int n;
    scanf("%d", &n);
    getchar();
    while (n--) {
        gets(s);
        int l = strlen(s);
        if (l == 0) {
            ++n;
            continue;
        }
        memset(ans, 0, sizeof(ans));
		memset(path, 0, sizeof(path));
        process(s[0]);
        for (i = 1; i < l; ++i) {
            process1(s[i], i);
        }        
        double max = -1;
        int index = 0;
        /*
        for (i = 0; i < 26; ++i)
            printf("%4c ", i + 'a');
        printf("\n");
        for (i = 0; i < l; ++i) {
            for (j = 0; j < 26; ++j) {
                printf("%lf ", ans[i][j]);
            }
            printf("\n");
        }
        */
        j = l - 1;
        char tt[1000];
        int xxx = 0;
        for (i = 0; i < 26; ++i) {
            if (ans[j][i] > max && check(i, s[j])) {
                max = ans[j][i];
                index = i;
            }
        }
        tt[xxx++] = index + 'a';
		int ta = index;
		i = l - 1;
		while (i) {
			tt[xxx++] = path[i][index] + 'a';
			index = path[i][index];
			--i;
		}
        for (i = xxx - 1; i >= 0; --i)
            printf("%c", tt[i]);
        printf("\n");
    }
    return 0;
}