HDU 1005 Wooden Sticks

Posted 2021-01-08 zlrrrr

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;
}