UVA 10128 - Queue(dp)

Queue

There is a queue with N people. Every person has a different heigth. We can see P people, when we are looking from the beginning, and R people, when we are looking from the end.It’s because they are having different height and they are covering each other.
How many different permutations of our queue has such a interesting feature?

Input Specification

The input consists of T test cases. The number of them (1<=T<=10000) is given on the first line of the input file.

Each test case consists of a line containing three integer numbers: N that indicates the number of people in a queue (1 <= N <= 13), and then two more integers. The first corresponds to the number of people, that we can see looking
from the beginning, and the second corresponding to the number of people, that we can see looking from the end.

Output Specification

For every test case your program has to determine one integer. Print how many permutations of N people we can see exactly P people from the beginning, and R people, when we are looking from the end.

Sample Input

3

10 4 4

11 3 1

3 1 2

Output for the Sample Input

90720

1026576

1

题意：给定n个人，每个人高度不同，然后给定两个数字，代表从前面能看到的人和从后面看能看到的人（高的会把矮的挡住）

思路： dp[i][j][k] 代表 i个人，从前面看j个，从后面看k个， 那么每次插入最矮的人，如果插在队头，j + 1, 如果插在队尾k + 1，插在中间是没有影响的。中间的位置有i - 1个，由此一来可以推出状态转移方程dp[i][j][k] = dp[i - 1][j - 1][k] + dp[i - 1][j][k - 1] + dp[i - 1][j][k] * (i - 2);

代码:

#include <stdio.h>
#include <string.h>
const int N = 15;

int t, n, q, h, dp

;

void init() {
dp[1][1][1] = 1;
for (int i = 2; i <= 13; i ++) {
for (int j = 1; j <= 13; j ++) {
for (int k = 1; k <= 13; k ++) {
dp[i][j][k] += dp[i - 1][j - 1][k] + dp[i - 1][j][k - 1] + dp[i - 1][j][k] * (i - 2);
}
}
}
}

int main() {
init();
scanf("%d", &t);
while (t --) {
scanf("%d%d%d", &n, &q, &h);
printf("%d\n", dp
[q][h]);
}
return 0;
}