hdu 1051 Wooden Sticks

问题描述

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 ) .

输入

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.

输出

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

样例输入

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

样例输出

2
1
3

先按照其中一个指标进行排序，然后再筛！！（从第一个开始，往后如果另一个指标都比这个大，那么就给个标记（也就是筛去了）。然后标记完了，又从第二个开始，筛过了就不用了，直接跳过，没有筛过就ans++。意思就是还有以这个打头的序列。）

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

struct stick
{
int l,w;
bool k;
}st[5005];

bool cmp(stick a,stick b)
{
return a.l!=b.l?a.l<b.l:a.w<b.w;
}

int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int k,ans=0;
scanf("%d",&k);
for(int i=1;i<=k;i++)
{
scanf("%d%d",&st[i].l,&st[i].w);
st[i].k=false;
}

sort(st+1,st+1+k,cmp);
for(int i=1;i<=k;i++)
{
if(!st[i].k)
{
st[i].k=true;
ans++;
int w=st[i].w;
for(int j=i+1;j<=k;j++)
{
if(!st[j].k&&st[j].w>=w)
{
st[j].k=true;
w=st[j].w;
}

}
}
}
printf("%d\n",ans);

}
}