问题描述

C/C++杭电1501题Wooden sticks 求挑错

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 (49) (52) (21) (35) and (14) then the minimum setup time should be 2 minutes since there is a sequence of pairs (14) (35) (49) (21) (52).

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

#include
#include

using namespace std;

struct stick
{
int len;
int wei;
bool judge;
}s[10009];

int cmp(stick astick b)
{
if(a.len!=b.len)
return a.len else
return a.wei}
int main()
{
int Tnijtime;
cin>>T;
while(T--)
{
cin>>n;
for(i=0;i {
cin>>s[i].len>>s[i].wei;
s[i].judge=true;
}
sort(ss+ncmp);
time=0;
int flag_len=s[0].lenflag_wei=s[0].weinum;
for(i=0;i {
if(s[i].judge==false)continue;
for(j=inum=0;j {
if(s[j].len>=flag_len&&s[j].wei>=flag_wei)
{
s[j].judge=false;
num++;
}
}
if(num)time++;
for(j=i+1;j<n;j++)
{
if(s[j].judge==true)
{
flag_len=s[j].len;
flag_wei=s[j].wei;
break;
}
}
}
cout<<time<<endl;
}
return 0;
}

解决方案

http://blog.csdn.net/lishuhuakai/article/details/8117251