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贪心算法 WOODEN STICKS 实例代码

  • 时间:2021-04-14 01:43 编辑: 来源: 阅读:
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摘要:贪心算法 WOODEN STICKS 实例代码
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
[u]复制代码[/u] 代码如下:
#include<stdio.h>  #include<stdlib.h>  #include<string.h>  #define N 5000;  struct node  {      int l;      int w;      int flag;  }sticks[5000];  int cmp(const void *p,const void *q)  {      struct node *a = (struct node *)p;      struct node *b = (struct node *)q;      if(a->l > b->l) return 1;      else if(a->l < b->l) return -1;      else return a->w > b->w ? 1 : -1;  }  int main()  {      int t,n,cnt,cl,cw;      int i,j;      scanf("%d",&t);      while(t--)      {          memset(sticks,0,sizeof(sticks));          scanf("%d",&n);          for( i = 0; i < n; i++)              scanf("%d %d",&sticks[i].l,&sticks[i].w);          qsort(sticks,n,sizeof(sticks[0]),cmp);          sticks[0].flag = 1;          cl = sticks[0].l;          cw = sticks[0].w;          cnt = 1;          for( j = 1; j < n; j++)          {              for( i = j; i < n; i++)              {                  if(!sticks[i].flag&&sticks[i].l>=cl&&sticks[i].w>=cw)                  {                      cl = sticks[i].l;                      cw = sticks[i].w;                      sticks[i].flag = 1;                  }              }              i = 1;              while(sticks[i].flag)                 i++;              j = i;              if(j == n) break;              cl = sticks[j].l;              cw = sticks[j].w;              sticks[j].flag = 1;              cnt++;          }          printf("%d\n",cnt);      }      return 0;  }
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。cl和cw是第一根的长度和重量,依次比较后面的是不是比当前的cl,cw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cl,cw,再循环往后比较。直到所有的都处理了。
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