HDU1051Wooden Sticks 贪心

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6563    Accepted Submission(s): 2707

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

// 贪心算法---先排序---后选择第一个没有用过的木头一次向后找，用掉所有可以用掉的木头，然后返回第一个没用过的木头继续找
#include<cstdio>
#include<algorithm>
using namespace std;
struct stick
{
    int len;
    int wei;
}a[5001];
bool used[5001];

bool cmp(stick k1,stick k2)       //按照长度从小到大排序，若长度相同按照重量递增排序
{
    if(k1.len==k2.len)
        return k1.wei<k2.wei;
    else
        return k1.len<k2.len;
}

int main()
{
    int T,n,i,j,st,count;
    bool flag;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(i=0;i<n;++i)
            scanf("%d%d",&a[i].len,&a[i].wei);
        sort(a,a+n,cmp);
        memset(used,false,sizeof(used));
        used[0]=true;
        st=0;       //记录第一个没有用过的木头
        count=0;
        while(st<n)
        {
            ++count;
            for(i=st+1,j=st,flag=true;i<n;++i)
            {
                if(used[i])
                    continue;
                if(a[j].len<=a[i].len&&a[j].wei<=a[i].wei)
                {
                    used[i]=true;
                    j=i;
                }
                else 
                {
                    if(flag)
                    {
                        st=i;      //只记录第一个没用过的木头
                        flag=false;
                    }
                }
            }
            if(flag)    //说明都用过了
                break;
        }
        printf("%d\n",count);
    }
    return 0;
}