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
简单贪心......
#include<iostream>
#include<algorithm>
using namespace std;
int t, n;
struct node
{
int l;
int w;
};
int cmp1(node a, node b)
{
if (a.l == b.l)
return a.w > b.w;
else
return a.l > b.l;
}
int main()
{
cin >> t;
while (t--)
{
cin >> n;
node box[5010], ans[5010];
for (int i = 1; i <= n; i++)
cin >> box[i].l >> box[i].w;
sort(box + 1, box + n + 1, cmp1);
int k = 1;
ans[1] = box[1];
for (int i = 2; i <= n; i++)
{
int flag = 0;
for (int j = 1; j <= k; j++)
{
if (box[i].l <= ans[j].l&&box[i].w <= ans[j].w)
{
ans[j].l = box[i].l, ans[j].w = box[i].w;
flag = 1;
break;
}
}
if (!flag)
ans[++k] = box[i];
}
if (n == 0)
cout << '0' << endl;
else
cout << k<< endl;
}
return 0;
}再附上JAVA版：
import java.util.*;
class node implements Comparable
{
public int l;
public int w;
node(int x1,int x2) //初始构造方法
{
this.l=x1;
this.w=x2;
}
public int compareTo(Object o)
{
node n=(node)o;
if(l==n.l)
{
if(w<n.w)
return 1;
else if(w>n.w)
return -1;
else
return 0;
}
else
{
if(l<n.l)
return 1;
else
return -1;
}
}
}
public class Main {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
for(int i=0;i<t;i++)
{
int num=0;
node ans[]=new node[5001];
int n=sc.nextInt();
node box[]=new node[5001];
for(int j=1;j<=n;j++) {
int tmp1=sc.nextInt(),tmp2=sc.nextInt();
box[j]=new node(tmp1,tmp2); //对于box的赋值需要使用其构造方法
}
Arrays.sort(box,1,n+1);
for(int j=1;j<=n;j++)
{
int flag=0;
for(int k=0;k<num;k++) {
if(box[j].l<=ans[k].l&&box[j].w<=ans[k].w)
{
ans[k].l=box[j].l;
ans[k].w=box[j].w;
flag=1;
break;
}
}
if(flag==0)
{
ans[num++]=new node(box[j].l,box[j].w); //同样对于ans的赋值也需要使用其构造方法
}
}
System.out.println(num);
}
}
}