HDOJ 1051

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8264    Accepted Submission(s): 3355

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

 

这道题是个比较好想的贪心 ， 按长度 ， 质量排序 ， 从第一个点开始，从区间上删除该点可以不费时间直接到达的一串儿点。

所费时间就是无法合并的集合数 ， 即有多少个起始点。

#include <iostream>#include <cmath>#include <cstdio>#include <algorithm>using namespace std;struct wood{    int l , w;}a[5005];bool f[5005];//记录这个位置是否取过bool cmp(wood a , wood b){    if(a.l != b.l)return a.l < b.l;    return a.w < b.w;}int main(){    int T , n;    cin >> T;    while(T--)    {        int time = 0;        cin >> n;        for(int i = 0 ; i < n ; i++)        {            scanf("%d%d",&a[i].l , &a[i].w);            f[i] = 0;        }        sort(a , a + n , cmp);//以长度为第一元素w为第二元素排序        for(int i = 0; i < n  ; i++)        {            if(f[i])continue;//如果这点取过，时间不增加            int t = a[i].w;            for(int j = i + 1 ; j < n ; j++)            {                if(!f[j]&&t <= a[j].w)                {                    t = a[j].w;                    f[j] = 1;                }            }            time ++;        }        cout << time << endl;    }    return 0;}