hdu 1051:Wooden Sticks（水题，贪心）

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

Asia 2001, Taejon (South Korea)

 

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　　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[5010];
int main()
{
    int T;
    cin>>T;
    while(T--){
        int n;
        cin>>n;
        for(int i=1;i<=n;i++)
            cin>>s[i].l>>s[i].w;
        //对l排序
        for(int i=1;i<=n-1;i++)
            for(int j=1;j<=n-i;j++){
                if(s[j].l>s[j+1].l){
                    int t;
                    t=s[j].l;s[j].l=s[j+1].l;s[j+1].l=t;
                    t=s[j].w;s[j].w=s[j+1].w;s[j+1].w=t;
                }
            }
        int sum=0;  //递增序列的个数
        //对w序列进行筛选
        while(1){
            int i;
            for(i=1;i<=n;i++){  //如果全部为-1，则退出循环
                if(s[i].w!=-1)
                    break;
            }
            if(i>n) break;
            sum++;
            int num=0;
            for(i=1;i<=n;i++){
                if(num==0 && s[i].w!=-1){
                    num=s[i].w;
                    s[i].w=-1;
                }
                else if(s[i].w!=-1 && s[i].w>=num){
                    num=s[i].w;
                    s[i].w=-1;
                }
            }
        }
        cout<<sum<<endl;

    }
    return 0;
}

Freecode : www.cnblogs.com/yym2013