ZOJ Problem Set - 1025

Wooden Sticks

Time Limit: 1 Second Memory Limit: 32768 KB

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 (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 n 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 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


Output for the Sample Input

2
1
3

#include <stdio.h>
#include <iostream>
using namespace std;

int s[5000][2];

void sort(int N)
{
int i, j, t[2];
for(i=1; i<N; i++)
{
t[0] = s[i][0];
t[1] = s[i][1];
for(j=i-1; j>=0; j--)
{
if(t[0] < s[j][0] || t[0] == s[j][0] && t[1] < s[j][1])
{
s[j+1][0] = s[j][0];
s[j+1][1] = s[j][1];
}
else break;
}
s[j+1][0] = t[0];
s[j+1][1] = t[1];
}
}

int main()
{
int T;
cin >> T;
for(int t=0; t<T; t++)
{
int N;
cin >> N;
for(int n=0; n<N; n++)
cin >> s
[0] >> s
[1];
sort(N);
int count = 0;
while(1)
{
int i = 0;
while(i<N && s[i][0] == -1) i++;
if(i==N) break;
int cur[2];
cur[0] = s[i][0];
cur[1] = s[i][1];
s[i][0] = -1;
for(; i<N; i++)
{
if(s[i][0] != -1 && s[i][0] >= cur[0] && s[i][1] >= cur[1])
{
cur[0] = s[i][0];
cur[1] = s[i][1];
s[i][0] = -1;
}
}
count ++;
}
cout << count << endl;
}
return 0;
}

这题用贪心算法可以实现全局最优。排个序，然后就能看出，每次找出的子序列都是无争议的最终情况。

搞个循环，把所有子序列都找出即可。