Hdoj1584 spider brand interval DP

Hdoj1584 spider brand interval DP

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Analysis:
F [I] [j] indicates the minimum number of steps required to stack a deck of cards I to j.
D [I] [j] indicates the number of steps required to move the card I to card j (in the initial status ).
Take a string of cards 3 ~ 8. For example, we need to place card 3 on Card 4. In the optimal mobile solution, the position of Card 4 is unknown, therefore, we can enumerate the position of Card 4 (because there are 10 cards in total, enumeration is acceptable) to obtain the state transition equation: f [3] [8] = min (f [3] [8], f [4] [k] + f [k] [8] + d [3] [k]); (4 <= k <= 8 ). F [I] [j] = min (f [I] [j], f [I + 1] [k] + f [k] [j] + d [I] [k]);

For example, the initial order of a card is 1, 4, 6, 8, 3, 2, 5, 7, 9, 10.
F [1] [4. In the most order: first move Card 2 to card 3, and then move Card 2 ~ 3 move to Card 4. Finally, Move card 1 to Card 2. Now Card 2 is in the position of Card 4. (F [1] [4] = f [2] [4] + f [4] [4] + d [1] [4] = 5 ).

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        Using namespace std; int t, a [15], d [15] [15], f [15] [15]; void dp () {for (int I = 0; I <10; I ++) // The obtained range is increasing. First, the interval with a small distance is obtained to obtain the optimal subproblem solution {for (int j = 1; j <= 10; j ++) // the minimum number of steps required to stack a string of cards from j to I + j. {If (I + j> 10) continue; for (int k = j + 1; k <= I + j; k ++) // The Position of the card on the enumeration. f [j] [I + j] = min (f [j] [I + j], f [j + 1] [k] + f [k] [I + j] + d [j] [k]) ;}} void Init () {for (int I = 1; I <= 10; I ++) scanf ("% d", & a [I]); memset (d, 0, sizeof (d); for (int I = 1; I <= 10; I ++) // preprocess all distances to {for (int j = 1; j <= 10; j ++) {int x = a [I], y = a [j]; d [x] [y] = abs (I-j ); d [y] [x] = d [x] [y] ;}} int main () {cin> t; while (t --) {for (int I = 1; I <= 10; I ++) {for (int j = 1; j <= 10; j ++) {f [I] [j] = 10e8; if (I = j) f [I] [j] = 0 ;}} Init (); dp (); printf ("% d \ n", f [1] [10]);} return 0 ;}