Wooden Sticks
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10423 Accepted Submission(s): 4287
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
Source
Recommend
水题。
先对木材长度 l 排序，在对 w 依次处理，将第一个递增序列筛选出来，筛选出来的元素赋值为-1，再将第二个递增序列筛选出来，同样标记-1 ，直到w序列全部为-1，代表筛选完成。输出记录下的递增序列的个数就是结果。
还可以用贪心做，有时间再做一遍。
代码：
#include <iostream> using namespace std;
struct stick{
int l,w;
}s[];
int main()
{
int T;
cin>>T;
while(T--){
int n;
cin>>n;
for(int i=;i<=n;i++)
cin>>s[i].l>>s[i].w;
//对l排序
for(int i=;i<=n-;i++)
for(int j=;j<=n-i;j++){
if(s[j].l>s[j+].l){
int t;
t=s[j].l;s[j].l=s[j+].l;s[j+].l=t;
t=s[j].w;s[j].w=s[j+].w;s[j+].w=t;
}
}
int sum=; //递增序列的个数
//对w序列进行筛选
while(){
int i;
for(i=;i<=n;i++){ //如果全部为-1，则退出循环
if(s[i].w!=-)
break;
}
if(i>n) break;
sum++;
int num=;
for(i=;i<=n;i++){
if(num== && s[i].w!=-){
num=s[i].w;
s[i].w=-;
}
else if(s[i].w!=- && s[i].w>=num){
num=s[i].w;
s[i].w=-;
}
}
}
cout<<sum<<endl; }
return ;
}
Freecode : www.cnblogs.com/yym2013