HDOJ 1051 Wooden Sticks

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15883    Accepted Submission(s): 6524

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

213

 

大致题意：

         工厂加工一批木棍，加工所用的设备调整一次需要一分钟，只要后面的木棍的长度和重量大于等于前面的木棍，设备就不用调整。设备第一次启动的时候也需要一分钟。木棍的顺序可以调整，依次输入木棍的数量，长度和重量，求加工完这些木棍所需的最短时间。

AC一：

思路：
因为题目中没说输入的木材是按长度从小到大（从大到小）排的，或是按重量从小到大（从大到小）排的，所以要先对木材进行排序，笔者是按木材长度从小到大排的（如果长度相同，则质量小的排前面）；然后就可以比较了，先定义两个变量来存放前一个木材的规格，再定义一个数组flag，1表示该木材已经加工，0反之，然后与后一个作比较，情况如下（注意，还得定义一个变量ans来记录已加工的木材的总数）：
        ①若符合不花费时间的标准，就加工（即把对应的vis标为1）；
        ②若不符合标准就跳过，等待下一轮的加工；
      

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 5000 + 10;struct P{    int l, w;}p[maxn];int t, n;int ans;bool flag[maxn];int cmp(P a, P b){    return a.l < b.l;}int main(){    scanf("%d", &t);    while (t--){        scanf("%d", &n);        for (int i = 0; i < n; i++){            scanf("%d%d", &p[i].l, &p[i].w);        }        sort(p, p + n, cmp);        ans = 0;        memset(flag, 0, sizeof(flag));        for (int i = 0; i < n; i++){            if (flag[i])                continue;            int L = p[i].l, W = p[i].w;            ans++;            for (int j = i; j < n; j++){                if (p[j].l >= L && p[j].w >= W && !flag[j]){                    L = p[j].l;                    W = p[j].w;                    flag[j] = 1;                }            }        }        printf("%d\n", ans);    }    return 0;}

AC二：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 5000 + 10;struct P{    int l, w;}p[maxn];int t, n;bool flag[maxn];int cmp(P a, P b){    if (a.l < b.l)        return 1;    else if (a.l == b.l && a.w < b.w)        return 1;    else        return 0;}int main(){    scanf("%d", &t);    while (t--){        scanf("%d", &n);        for (int i = 0; i < n; i++){            scanf("%d%d", &p[i].l, &p[i].w);        }        sort(p, p + n, cmp);        int ans = 0;        for (int i = 0; i < n; i++){            if (p[i].w){                ans++;                int temp = p[i].w;                p[i].w = 0;                for (int j = i + 1; j < n; j++){                    if (p[j].w >= temp){                        temp = p[j].w;                        p[j].w = 0;                    }                }            }        }        printf("%d\n", ans);    }    return 0;}