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  • 贪心算法 Wooden Sticks

     

    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
     
     1 #include<stdio.h>
     2 #include<stdlib.h>
     3 #include<string.h>
     4 #define N 5000;
     5 
     6 struct node
     7 {
     8     int l;
     9     int w;
    10     int flag;
    11 }sticks[5000];
    12 int cmp(const void *p,const void *q)
    13 {
    14     struct node *a = (struct node *)p;
    15     struct node *b = (struct node *)q;
    16     if(a->l > b->l) return 1;
    17     else if(a->l < b->l) return -1;
    18     else return a->w > b->w ? 1 : -1;
    19 }
    20 int main()
    21 {
    22     int t,n,cnt,cl,cw;
    23     int i,j;
    24     scanf("%d",&t);
    25     while(t--)
    26     {
    27         memset(sticks,0,sizeof(sticks));
    28         scanf("%d",&n);
    29         for( i = 0; i < n; i++)
    30             scanf("%d %d",&sticks[i].l,&sticks[i].w);
    31         qsort(sticks,n,sizeof(sticks[0]),cmp);
    32         sticks[0].flag = 1;
    33         cl = sticks[0].l;
    34         cw = sticks[0].w;
    35         cnt = 1;
    36         for( j = 1; j < n; j++)
    37         {
    38             for( i = j; i < n; i++)
    39             {
    40                 if(!sticks[i].flag&&sticks[i].l>=cl&&sticks[i].w>=cw)
    41                 {
    42                     cl = sticks[i].l;
    43                     cw = sticks[i].w;
    44                     sticks[i].flag = 1;
    45                 }
    46             }
    47             i = 1;
    48             while(sticks[i].flag)
    49                i++;
    50             j = i;
    51             if(j == n) break;
    52             cl = sticks[j].l;
    53             cw = sticks[j].w;
    54             sticks[j].flag = 1;
    55             cnt++;
    56         }
    57         printf("%d\n",cnt);
    58 
    59     }
    60     return 0;
    61 }

    题意:

    我们要处理一些木棍,第一根的时间是1分钟,另外的,在长度为l,重为w的木棍后面的那根的长度为l’, 重量w’,只要l <=l’ 并且w <= w’,就不需要时间,否则需要1分钟,求如何安排处理木棍的顺序,才能使花的时间最少。

    思路:

    贪心算法。先把这些木棍按长度和重量都从小到大的顺序排列。clcw是第一根的长度和重量,依次比较后面的是不是比当前的clcw大,是的话把标志flag设为1,并跟新cl,cw。比较完后,再从前往后扫描,找到第一个标志位为0的,作为是第二批的最小的一根,计数器加一。把它的长度和重量作为当前的clcw,再循环往后比较。直到所有的都处理了。

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  • 原文地址:https://www.cnblogs.com/LK1994/p/3099538.html
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