HDU 6076 题解

题目：http://acm.hdu.edu.cn/showproblem.php?pid=6076

Problem Description

In airport of Bytetown, there are two long queues waiting for security check. Checking a person needs one minute, and two queues can be checked at the same time.

Two teams A and B are going to travel by plane. Each team has n players, ranked from 1 to n according to their average performance. No two players in the same team share the same rank. Team A is waiting in queue 1 while team B is waiting in queue 2. Nobody else is waiting for security check.

Little Q is the policeman who manages two queues. Every time he can check one person from one queue, or check one each person from both queues at the same time. He can’t change the order of the queue, because that will make someone unhappy. Besides, if two players Ai and Bj are being checked at the same time, satisfying |Ai−Bj|≤k, they will make a lot of noise because their rank are almost the same. Little Q should never let that happen.

Please write a program to help Little Q find the best way costing the minimum time.

Input
The first line of the input contains an integer T(1≤T≤15), denoting the number of test cases.

In each test case, there are 2 integers n,k(1≤n≤60000,1≤k≤10) in the first line, denoting the number of players in a team and the parameter k.

In the next line, there are n distinct integers A1,A2,...,An(1≤Ai≤n), denoting the queue 1 from front to rear.

Then in the next line, there are n distinct integers B1,B2,...,Bn(1≤Bi≤n), denoting the queue 2 from front to rear.

Output

For each test case, print a single line containing an integer, denoting the minimum time to check all people.

Sample Input

14 22 3 1 41 2 4 3

Sample Output

7HintTime 1 : Check A_1.Time 2 : Check A_2.Time 3 : Check A_3.Time 4 : Check A_4 and B_1.Time 5 : Check B_2.Time 6 : Check B_3.Time 7 : Check B_4.

代码：

/*HDU 6076 - Security Check [ DP,二分 ]  |  2017 Multi-University Training Contest 4题意：    给出两个检票序列 A[N], B[N]    规定 abs(A[i]-B[j]) <= k 的i,j不能同时检票    求最少的检票时间    限制  N<= 6e4, k <= 10分析：    f(i,j) 为检票至i,j的时间    则 f(i,j) = f(i-1,j-1) + 1 ,  abs(A[i]-B[j]) > k              = min(f(i-1,j), f(i,j-1)) + 1 , abs(A[i]-B[j]) <= k    对于第二项，由于k小，可DP    对于第一项，可以二分最大的 t 使得 f(i,j) = f(i-t,j-t) + t 成立        那么 f(i-t,j-t) 就是第二项的了*/ #include <bits/stdc++.h>using namespace std;const int N = 60005;vector<int> G[N<<1];int dp[N][25];int a[N], b[N], pos[N];int t, n, k;int f(int n, int m){    if (n == 0 || m == 0) return n+m;    if (abs(a[n]-b[m]) > k)    {        int t = lower_bound(G[m-n+N].begin(), G[m-n+N].end(), n) - G[m-n+N].begin();        if (t == 0) return max(n, m);        t = G[m-n+N][t-1];        return f(t, m-n+t) + n-t;    }    else    {        int t = b[m]-a[n]+k;        if (dp[m][t] == -1)            dp[m][t] = min(f(n-1, m), f(n, m-1)) + 1;        return dp[m][t];    }}int main(){    scanf("%d", &t);    while (t--)    {        for (int i = 0; i < N<<1; i++) G[i].clear();        memset(dp, -1, sizeof(dp));        scanf("%d%d", &n, &k);        for (int i = 1; i <= n; i++)        {            scanf("%d", &a[i]);            pos[a[i]] = i;        }        for (int i = 1; i <= n; i++) scanf("%d", &b[i]);        for (int i = 1; i <= n; i++)            for (int j = max(1, b[i]-k); j <= min(n, b[i]+k); j++)            {                G[i-pos[j]+N].push_back(pos[j]);// 记录当a、b保持当前相对位置（i-pos[j]+N）时，a的哪些位置是不能同时检查的            }        for (int i = 0; i < N<<1; i++) sort(G[i].begin(), G[i].end());        int ans = f(n, n);        printf("%d\n", ans);    }}

转载文章：HDU 6076 - Security Check | 2017 Multi-University Training Contest 4