UVAlive 2322 Wooden Sticks(贪心)

There is a pile of 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:

  • The setup time for the first wooden stick is 1 minute.
  • Right after processing a stick of length and weight , the machine will need no setup time for a stick of length l' and weight w' if <=l' and <=w' .

Otherwise, it will need 1 minute for setup.
You are to find the minimum setup time to process a given pile of  wooden sticks. For example, if you have five sticks whose pairs of length and weight are (4,9), (5,2), (2,1), (3,5), and (1,4) then the minimum setup time should be 2 minutes since there is a sequence of pairs (1,4), (3,5), (4,9), (2,1), (5,2).

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 2 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  th wooden stick, respectively. The 2 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

题意: 给定n个木棍,每个木棍有重量w和长度l。现在要加工木棍。每次加工木棍要设置一下,如果下次加工的木棍重量长度w'、l‘和当前木棍加工长度w、l满足条件。w'>=w 和 l'>=l就不需要重新设置。求最少设置次数。

 

思路:贪心。把木棍按w从小到大排序。相同的按l从小到大排序。然后开一个标记数组。已经加工过的标记掉。每次从开头找到一根未加工的。去寻求最多不用重新设置可以加工掉的木棍。全部加工掉。直到全部加工掉。

代码:

 

#include <stdio.h>

#include <string.h>

#include <algorithm>

using namespace std;



int t, n, time;

struct Strick {

    int l, w, v;//v用来标记

} s[5005];



int cmp(Strick a, Strick b) {

    if (a.l != b.l)

	return a.l < b.l;

    return a.w < b.w;

}



int main() {

    scanf("%d", &t);

    while (t --) {

	time = 0;

	memset(s, 0, sizeof(s));

	scanf("%d", &n);

	for (int i = 0; i < n; i ++)

	    scanf("%d%d", &s[i].l, &s[i].w);

	sort(s, s + n, cmp);

	for (int i = 0; i < n; i ++) {

	    if (s[i].v) continue;

	    int sb = s[i].w;

	    time ++;

	    for (int j = i; j < n; j ++)

		if (s[j].w >= sb && !s[j].v) {

		    sb = s[j].w;

		    s[j].v = 1;

		}

	}

	printf("%d\n", time);

    }

    return 0;

}


 


 

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