HDU1051 Wooden Sticks 贪心

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11906 Accepted Submission(s): 4943


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
 
翻译一下题目:
现有n根木棒,已知它们的长度和重量。要用一部木工机一根一根地加工这些木棒。该机器在加工过程中需要一定的准备时间,是用于清洗机器,调整工具和模板的。木工机需要的准备时间如下:
(1)第一根木棒需要1min的准备时间;
(2)在加工了一根长为l,重为w的木棒之后,接着加工一根长为ll(l<=ll),重为ww(w<=ww)的木棒是不需要任何准备时间的。否则需要一分钟的准备时间。
给定n根木棒,你要找到最少的准备时间。例如现在有长和重分别为(4,9),(5,2),(2,1),(3,5)和(1,4)的五根木棒,那么所需准备时间最少为2min,顺序为(1,4),(3,5),(4,9),(2,1),(5,2)。
 
简单的贪心,只需要不停的扫描这n个木棍,符合条件就继续往后扫,已经比过的直接跳过,到最后没了时间就+1,然后再从头扫描,用一个cnt来计数,循环结束的条件是cnt==n。  注意一定要先排序!!
 
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;

struct stick
{
    int l,w;
    int v;
}s[5500];

int cmp(const stick &a,const stick &b)
{
    if(a.l==b.l)
        return a.w<b.w;
    return a.l<b.l;
}

int main()
{
    int i,t,j,cnt,k,ans,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            scanf("%d%d",&s[i].l,&s[i].w);
            s[i].v=0;
        }
        sort(s,s+n,cmp);
        cnt=0;k=0;ans=0;
        while(cnt!=n)
        {
            k=0;
            while(k<n && s[k].v==1)
                k++;
            ans++;
            for(i=k;i<n;i++)
            {
                if(s[i].v==0 && s[i].l>=s[k].l && s[i].w>=s[k].w)
                {
                    s[i].v=1;cnt++;
                    k=i;
                }
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}

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