HDU1051 Wooden Sticks

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16395    Accepted Submission(s): 6731

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)

題意:機器處理木材，每次開始時需要1分鐘的設置是時間，如果下一根木棒的長度和重量都大於或等於上一根木棒，呢麼就不用去設置，請問怎樣設置才能使化肥設置及其的時間最短。

這是剛學貪心算法，按照書上的運行了一遍，弄懂了

思路：將木棒進行自定義排序，長度小，重量輕的排在前邊，然後每次從後邊的比較，然後算出來最小時間！

代碼如下“：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define MAXN 5005
using namespace std;

struct Node
{
    int length,weight;
    bool flag;    //  flag用來標記這根木棒是否處理過
};

bool cmp(Node a,Node b)
{
    return a.length<=b.length || (a.length==b.length && a.weight<=b.weight);   //自定義排序，長度小，質量輕的排在前邊
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        Node node[MAXN];
        int n,cot=0,k;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&node[i].length,&node[i].weight);
            node[i].flag=false;
        }
        sort(node,node+n,cmp); //自定義排序，當然有些條件不滿足的，可以輸出看一下
        for(int i=0;i<n;i++)
        {
            if(node[i].flag) continue;  //如果這跟木棒處理過，直接進行下一個循環，知道找到新的未處理的木棒
            cot++;k=i;   //這時候，花費的時間+1
            for(int j=i+1;j<n;j++)
            {
                if(node[j].length>=node[k].length && node[j].weight>=node[k].weight && node[j].flag==false) // 如果這跟木棒長度，質量都大於上一根而且沒有處理過
                {
                    node[j].flag=true;  //處理這一跟木棒
                    k=j;//更新一下，把這一跟作爲下一次的對比，看下一根木棒的長度，質量是否滿足條件
                }
            }
        }
        printf("%d\n",cot);
    }
    return 0;
}

總結：貪心，每次要看出能貪心的對象，才能去做