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貪心算法 WOODEN STICKS 實(shí)例代碼

更新時(shí)間：2013年05月26日 11:13:49 作者：

貪心算法 WOODEN STICKS 實(shí)例代碼，需要的朋友可以參考一下

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

復(fù)制代碼代碼如下:

#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;
 }

題意：

我們要處理一些木棍，第一根的時(shí)間是1分鐘，另外的，在長(zhǎng)度為l，重為w的木棍后面的那根的長(zhǎng)度為l', 重量w'，只要l <=l' 并且w <= w'，就不需要時(shí)間，否則需要1分鐘，求如何安排處理木棍的順序，才能使花的時(shí)間最少。

思路：

貪心算法。先把這些木棍按長(zhǎng)度和重量都從小到大的順序排列。cl和cw是第一根的長(zhǎng)度和重量，依次比較后面的是不是比當(dāng)前的cl，cw大，是的話把標(biāo)志flag設(shè)為1，并跟新cl,cw。比較完后，再?gòu)那巴髵呙?，找到第一個(gè)標(biāo)志位為0的，作為是第二批的最小的一根，計(jì)數(shù)器加一。把它的長(zhǎng)度和重量作為當(dāng)前的cl，cw，再循環(huán)往后比較。直到所有的都處理了。

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