UVa 1638 Pole Arrangement (recursive or DP)

Test instructions: There is a high for the 1,2,3...N pole each row, from the left can see L root, from the right can see R root, to find out how many kinds of pole arrangement is possible.

Analysis: Set D (i, J, K) to indicate that a pole with a height of 1-i is lined up, see J Root from the left, and see the number of K roots from the right. When I>1, we press the pole in order from big to small,

Assuming that the i-1 root has been arranged, then there is a height of 1, then it will not block any pole. There are 3 types of cases:

1. On the left, you will be able to see it on the left and not see it on the right (because i>1);

2. On the right side, you will be able to see it on the right, not on the left (because I>1);

3. The rest is put in the middle, then we have how many places can be put, the answer is obvious, I-2, can not see it on either side.

Situation 1 o'clock, in the height in the 2-i pole, on the left can see J-1 root, from the right can see K root, because on the left with the first root is just J root,

Similarly, the case 2, in the height of the 2-i pole, on the left can see J Root, from the right can see k-1 root, because on the right with the first root is just k root,

What about Scenario 3? Better said, in the height of the pole in the 2-i, on the left can see J Root, from the right can see K root, because on the left and right are not see the first root.

Comprehensive analysis: D (i, j, k) = d (I-1, j-1, K) + D (I-1, J, K-1) + D (I-1, J, K) * (i-2), boundary is D (1, 1, 1) = 1;

You can first sweep out the situation first, when the direct query on the line, fast and convenient. Note that more than int is used to store a long long.

The code is as follows:

#include <iostream> #include <cstdio> #include <cstring> #define MoD%10056using namespace Std;typedef Long Long LL; LL d[25][25][25];void init () {    memset (d, 0, sizeof (d));    D[1][1][1] = 1;    for (int i = 2; I <=, ++i) for        (int j = 1; J <=, ++j) for            (int k = 1; k <=; ++k)                d[i][j][k] = d [I-1] [J-1] [K] + d[i-1][j][k-1] + d[i-1][j][k] * (i-2);} int main () {    init ();    int T, N, L, R;  Cin >> T;    while (t--) {        scanf ("%d%d%d", &n, &l, &r);        printf ("%lld\n", D[n][l][r]);    }    return 0;}

UVa 1638 Pole Arrangement (recursive or DP)