UVa10653.Prince and Princess

题目连接：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1576

13935381

10635

Prince and Princess

Accepted

C++

0.095

2014-07-24 03:41:18

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Prince and Princess
Input: Standard Input

Output: Standard Output

Time Limit: 3 Seconds

In an n x n chessboard, Prince and Princess plays a game. The squares in the chessboard are numbered 1, 2, 3 ... n*n, as shown below:

Prince stands in square 1, make p jumps and finally reach square n*n. He enters a square at most once. So if we use x_p to denote the p-th square he enters, then x₁, x₂, ... x_p+1 are all different. Note that x₁ = 1 and x_p+1 = n*n. Princess does the similar thing - stands in square 1, make q jumps and finally reach square n*n. We use y₁, y₂ , ... y_q+1 to denote the sequence, and all q+1 numbers are different.

Figure 2 belows show a 3x3 square, a possible route for Prince and a different route for Princess.

The Prince moves along the sequence: 1 --> 7 --> 5 --> 4 --> 8 --> 3 --> 9 (Black arrows), while the Princess moves along this sequence: 1 --> 4 --> 3 --> 5 --> 6 --> 2 --> 8 --> 9 (White arrow).

The King -- their father, has just come. "Why move separately? You are brother and sister!" said the King, "Ignore some jumps and make sure that you're always together."

For example, if the Prince ignores his 2nd, 3rd, 6th jump, he'll follow the route: 1 --> 4 --> 8 --> 9. If the Princess ignores her 3rd, 4th, 5th, 6th jump, she'll follow the same route: 1 --> 4 --> 8 --> 9, (The common route is shown in figure 3) thus satisfies the King, shown above. The King wants to know the longest route they can move together, could you tell him?

Input

The first line of the input contains a single integer t(1 <= t <= 10), the number of test cases followed. For each case, the first line contains three integers n, p, q(2 <= n <= 250, 1 <= p, q < n*n). The second line contains p+1 different integers in the range [1..n*n], the sequence of the Prince. The third line contains q+1 different integers in the range [1..n*n], the sequence of the Princess.

Output

For each test case, print the case number and the length of longest route. Look at the output for sample input for details.

 

Sample Input         Output for Sample Input

1 3 6 7 1 7 5 4 8 3 9 1 4 3 5 6 2 8 9

Case 1: 4

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解题思路：由于跪在校赛的LCS nlogn算法，所以回来恶补，特意找了一道相同意思的题目。如同此题，简单的LCS类型，但是10^5左右的数据量，如果开滚动dp的话，复杂度O(n^2)会超时。所以应该将LCS问题转化成下标对应的LIS问题，然后再使用LIS的nlogn算法，具体思想是维护一个递增序列，二分找上界，替换元素即可。（感谢DIM神给我讲解此算法）。这样已经足够了。但是说句与此题无关的话，如果数据范围过大，例如到64位级别的数字时，可以使用离散化，继续降低复杂度。膜拜下yeahpeng酱。

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 #include <cstdlib>
 #include <cctype>
 #include <algorithm>
 #include <numeric>
 #include <map>
 #include <vector>
 using namespace std;

 const int N =  * ;
 map<int, int> check;
 vector<int> Stack;

 int main () {
     int T, n, p, q, cur = ;
     int s1[N], s2[N];
     scanf("%d", &T);
     while (T --) {
         Stack.clear();
         check.clear();
         scanf("%d%d%d", &n, &p, &q);
         for (int i = ; i <= p; ++ i) {
             scanf("%d", &s1[i]);
             check[s1[i] ] = i + ;
         }

         map<int, int> :: iterator it;

         for (int i = ; i <= q; ++ i) {
             scanf("%d", &s2[i]);
             if (check.find(s2[i]) != check.end()) {
                 s2[i] = check[s2[i]];
             } else {
                 s2[i] = ;
             }
         }
         /*
         for (int i = 0 ; i <= p; ++ i) {
             cout << check[s1[i] ] << " ";
         }
         cout << endl;
         *//*
         for (int i = 0 ; i <= q; ++ i) {
             cout << s2[i] << " ";
         }
         cout << endl;
 */
         for (int i = ; i <= q; ++ i) {
             if (Stack.empty() || s2[i] > Stack[Stack.size() - ]) {
                 Stack.push_back(s2[i]);
             } else {
                 vector<int> :: iterator it = lower_bound(Stack.begin(), Stack.end(), s2[i]);
                 *it = s2[i];
             }
         }
         /*for (int i =  0; i < Stack.size(); ++ i ) {
             cout << Stack[i] << " ";
         }
         cout << endl;*/
         printf("Case %d: ", ++ cur);
         cout << Stack.size() << endl;

     }
     return ;
 }