hdu_1051 Wooden Sticks 贪心

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 27724    Accepted Submission(s): 11221

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).

 

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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.

 

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Output

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

 

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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

 

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Sample Output

2

1

3

 

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按照长度 l 简单排一下序， 从头开始遍历， 每次找可能多的 比前一根木头重量小的。

#include <iostream>
#include <stdio.h>
#include <algorithm>

using namespace std;

struct dat
{
    int l;
    int w;
    int visit;
} data[];

bool cmp(dat a, dat b)
{
    return a.l<b.l;
}

int main()
{
    int t, n;
    scanf("%d",&t);

    while(t--)
    {
        scanf("%d", &n);
        for(int i=; i<n; i++)
        {
            scanf("%d%d",&data[i].l, &data[i].w);
            data[i].visit=;
        }

        sort(data, data+n, cmp);

        int temp, ans=;

        for(int i=; i<n; i++)
        {
            if(!data[i].visit)
            {
                ans++;
                data[i].visit=;
                temp = i;
                for(int j=i+; j<n; j++)
                {
                    if(!data[j].visit && data[temp].l<=data[j].l
                            && data[temp].w <= data[j].w)
                    {
                        temp = j;
                        data[j].visit = ;
                    }
                }
            }
        }

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

    }
    return ;
}