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    Wooden Sticks

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 11023    Accepted Submission(s): 4530


    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

    【题目来源】

    Asia 2001, Taejon (South Korea)

    【题目大意】
    有一堆木棒,这些木棒的长度和重量是给定的。这些木棒将被机器一根接一根的进行处理。
    机器处理每一根木棒需要一些时间,这个时间被叫做建立时间。
    这个时间是这样定义的:
    1.第一根木棒的建立时间是1分钟;
    2.在处理一根长度为l、重量为w的木棒之后,如果木棒的l'<=l、w<=w',那么这个机器将不需要建立时间。否则,这个机器将需要1分钟的建立时间。
    你需要找到处理完这些木头的最短建立时间。题目中的例子看懂就知道怎么做了。

    【题目分析】

    首先按照l排序,当l相同的时候,再按照w排序,然后就是按照顺序一遍一遍的标记并统计,直至所有的都被标记,最后的count就是答案。

    奇怪的是,我的代码在测试样例的时候是错的,我提交后竟然ac了。

    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    struct Node
    {
        int l,w;
        bool mark;
    };
    
    bool cmp(Node a,Node b)
    {
        if(a.l==b.l)
        {
            return a.w<b.w;
        }
        else return a.l<b.l;
    }
    
    Node node[5050];
    int main()
    {
        int T;
        scanf("%d",&T);
        while(T--)
        {
            int n,i,j;
            scanf("%d",&n);
            for(i=0;i<n;i++)
            {
                scanf("%d%d",&node[i].l,&node[i].w);
                node[i].mark=0;
            }
            sort(node,node+n,cmp);
            int cl=node[0].l,cw=node[0].w;
            node[0].mark=1;
            int cnt=1;
            for(i=1;i<n;i++)
            {
                for(j=i;j<n;j++)
                {
                    if(!node[j].mark&&node[j].l>=cl&&node[j].w>=cw)
                    {
                        node[j].mark=1;
                        cl=node[j].l;
                        cw=node[j].w;
                    }
                }
                j=1;
                while(node[j].mark)
                    j++;
                i=j;
                if(i==n)
                    break;
                cnt++;
                node[i].mark=1;
                cl=node[i].l;
                cw=node[i].w;
            }
            printf("%d
    ",cnt);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/crazyacking/p/3754177.html
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