Wooden Sticks
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 27724 Accepted Submission(s): 11221
Problem Description
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
Sample Output
2
1
3
按照长度 l 简单排一下序， 从头开始遍历， 每次找可能多的 比前一根木头重量小的。
#include <iostream>
#include <stdio.h>
#include <algorithm> using namespace std; struct dat
{
int l;
int w;
int visit;
} data[]; bool cmp(dat a, dat b)
{
return a.l<b.l;
} int main()
{
int t, n;
scanf("%d",&t); while(t--)
{
scanf("%d", &n);
for(int i=; i<n; i++)
{
scanf("%d%d",&data[i].l, &data[i].w);
data[i].visit=;
} sort(data, data+n, cmp); int temp, ans=; for(int i=; i<n; i++)
{
if(!data[i].visit)
{
ans++;
data[i].visit=;
temp = i;
for(int j=i+; j<n; j++)
{
if(!data[j].visit && data[temp].l<=data[j].l
&& data[temp].w <= data[j].w)
{
temp = j;
data[j].visit = ;
}
}
}
} printf("%d\n",ans); }
return ;
}