• HDU 1005 Wooden Sticks


    http://acm.hdu.edu.cn/showproblem.php?pid=1051

    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
     
    题解:贪心 先排序
    时间复杂度:$O(N ^ 2 + N * logN)$
    代码:
    #include <bits/stdc++.h>
    using namespace std;
    
    const int maxn = 1e5 + 10;
    int N;
    
    struct Node {
        int l;
        int w;
        int flag;
    }node[maxn];
    
    bool cmp(const Node& a, const Node& b) {
        return a.l == b.l ? a.w < b.w : a.l < b.l;
    }
    
    int main() {
        int T;
        scanf("%d", &T);
        while(T --) {
            memset(node, 0, sizeof(node));
            scanf("%d", &N);
            for(int i = 1; i <= N; i ++) {
                scanf("%d%d", &node[i].l, &node[i].w);
                node[i].flag = 0;
            }
    
            int cnt = 0;
            sort(node + 1, node + 1 + N, cmp);
            for (int k = 1; k <= N;)
            {
                cnt ++;
                int L = 0, W = 0;
                for (int i = 1; i <= N; i ++) {
                    if (!node[i].flag)
                    if (node[i].l >= L && node[i].w >= W) {
                        node[i].flag = 1;
                        L = node[i].l;
                        W = node[i].w;
                        k ++;
                    }
                }
            }
            printf("%d
    ", cnt);
        }
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/zlrrrr/p/9698709.html
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