Wooden Sticks POJ - 1065 (DP)

There is a pile of n wooden sticks. The length and weight of each stick are known in advance. The sticks are to be processed by a woodworking machine in one by one fashion. It needs some time, called setup time, for the machine to prepare processing a stick. The setup times are associated with cleaning operations and changing tools and shapes in the machine. The setup times of the woodworking machine are given as follows: 
(a) The setup time for the first wooden stick is 1 minute. 
(b) Right after processing a stick of length l and weight w , the machine will need no setup time for a stick of length l' and weight w' if l <= l' and w <= w'. Otherwise, it will need 1 minute for setup. 
You are to find the minimum setup time to process a given pile of n wooden sticks. For example, if you have five sticks whose pairs of length and weight are ( 9 , 4 ) , ( 2 , 5 ) , ( 1 , 2 ) , ( 5 , 3 ) , and ( 4 , 1 ) , then the minimum setup time should be 2 minutes since there is a sequence of pairs ( 4 , 1 ) , ( 5 , 3 ) , ( 9 , 4 ) , ( 1 , 2 ) , ( 2 , 5 ) .

Input

The input consists of T test cases. The number of test cases (T) is given in the first line of the input file. Each test case consists of two lines: The first line has an integer n , 1 <= n <= 5000 , that represents the number of wooden sticks in the test case, and the second line contains 2n positive integers l1 , w1 , l2 , w2 ,..., ln , wn , each of magnitude at most 10000 , where li and wi are the length and weight of the i th wooden stick, respectively. The 2n integers are delimited by one or more spaces.

Output

The output should contain the minimum setup time in minutes, one per line.

Sample Input

3 
5 
4 9 5 2 2 1 3 5 1 4 
3 
2 2 1 1 2 2 
3 
1 3 2 2 3 1 

Sample Output

2
1
3

解题思路:

先根据第一根筷子的长度从左到右排序,然后求最长下降序列(最长上升序列反一下)

#include <iostream>
#include <algorithm>
#include <string.h>
using namespace std;
const int MAXN = 5010;
int N,K;
int stDp[MAXN]; 
struct stick {
	int x, y;  //首,尾
}sticks[MAXN];
bool scmp(stick a, stick b) {
	if (a.x != b.x) {
		return a.x < b.x;
	}
	else {
		return a.y < b.y;
	}
}
int main() {
	cin >> N;
	while (N--) {
		cin >> K;
		for (int i = 0; i < K; ++i) {
			cin >> sticks[i].x >> sticks[i].y;
		}
		sort(sticks, sticks + K,scmp); //按照初始长度排序
		memset(stDp, 0, sizeof(stDp));
		for (int i = 0; i < K; ++i) {
			stDp[i] = 1;  //记住一定要赋初值
			for (int j = 0; j < i; ++j) {
				if (sticks[i].y < sticks[j].y) {
					stDp[i] = max(stDp[i], stDp[j] + 1);
				}
			}
		}
		cout <<*max_element(stDp, stDp + K) << endl;
	}

	system("PAUSE");
	return 0;
}