Wooden Sticks
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11694 Accepted Submission(s): 4837
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
题目大意:
给你n个木棒，每个木棒长度为l、重量为w，然后用机器加工木棒，机器加工第一个木棒时间为1，若加工完第i个木棒后，再加工第j个木棒，若l[j]>=l[i]&&w[j]>=w[i]则加工第j个木棒不需要消耗时间。求加工完n条木棒所需要的最短时间。
思路：
想做动归题目，看了一下HDU题目分类，但是没想出动归转移方程，用并查集和贪心水了一下就过了。。。= = 。。
这道题很明显使得木棒构成的链最长，一条链就是前面木棒i的l和w都小于等于后面木棒j的l和w, 当链最长的时候，那么链的个数最少，于是花费的时间也最短。怎么使得链最长呢，就需要贪心了，先对l从小到大排序，再按w从小到大排序，然后当第j个木棒的l和w都大于等于第i个木棒的l和w的话就把它们并到一个集合里。最后算一下几个集合即为时间。
代码：
1 #include <cstdio>
2 #include <algorithm>
3 #include <cstring>
4 #include <vector>
5 #include <iostream>
6 using namespace std;
7 #define N 5005
8
9 struct node{
10 int id;
11 int l, w;
12 int flag;
13 }a[N];
14
15 int father[N];
16 int findroot(int p){
17 int r=p;
18 while(r!=father[r]) r=father[r];
19 return r;
20 }
21
22 bool cmp(node a,node b){
23 if(a.l==b.l) return a.w<b.w;
24 return a.l<b.l;
25 }
26
27 main()
28 {
29 int t;
30 int n, i, j, ans, k;
31 cin>>t;
32 while(t--){
33 scanf("%d",&n);
34 for(i=0;i<=n;i++) father[i]=i;
35 k=0;
36 for(i=0;i<n;i++) {
37 scanf("%d %d",&a[i].l,&a[i].w);
38 a[i].id=k++;
39 a[i].flag=0;
40 }
41 sort(a,a+n,cmp);
42 for(i=0;i<n;i++){
43 if(!a[i].flag){
44 k=i;
45 for(j=i+1;j<n;j++){
46 if(!a[j].flag&&a[j].l>=a[k].l&&a[j].w>=a[k].w){
47 father[a[j].id]=a[k].id;
48 a[j].flag=1;
49 k=j;
50 }
51 }
52 }
53
54 }
55 ans=0;
56 for(i=0;i<n;i++) {
57 if(findroot(a[i].id)==a[i].id)
58 ans++;
59 }
60 printf("%d\n",ans);
61 }
62 }
HDU 1051 并查集+贪心,布布扣,bubuko.com