题目描述:
给定n个数对(xi,yi),对这些数对进行划分,保证划分出来的每一堆都是一条单调的链,问最少可以划分成多少堆
原题如下:
Wooden Sticks
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 26409 | Accepted: 11423 |
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 ( 9 , 4 ) , ( 2 , 5 ) , ( 1 , 2 ) , ( 5 , 3 ) , and ( 4 , 1 ) , then the minimum setup time should be 2 minutes since there is a sequence of pairs ( 4 , 1 ) , ( 5 , 3 ) , ( 9 , 4 ) , ( 1 , 2 ) , ( 2 , 5 ) .
(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 ( 9 , 4 ) , ( 2 , 5 ) , ( 1 , 2 ) , ( 5 , 3 ) , and ( 4 , 1 ) , then the minimum setup time should be 2 minutes since there is a sequence of pairs ( 4 , 1 ) , ( 5 , 3 ) , ( 9 , 4 ) , ( 1 , 2 ) , ( 2 , 5 ) .
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 2n 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
Source
poj1548原理相同,不放了
首先我们要求单调链,把样例中的点画在坐标系中,我们便可看出,要求出单调链,要对数据排序(二维降一维)
poj讨论区有较为清楚的描述(本人理解但是无法用语言描述)
第一次贪心:其实并非每次都要求一个最长的序列,然后删除,只要选出的序列符合"极大"即可。这里的极大和函数的极值是一个概念。 例如,(1,4) (4,4) (5,3) (9,4) 其中最大(最长)的一个序列为 (1,4)->(4,4)->(9,4) 极大的序列 为 (5,3)->(9,4) 证明:很简单,因为所有的木头都需要被加工,并且与加工的先后顺序无关,那么该木头一定 需要在某个序列中,而我们每次选择的序列是极大的,故每次操作都没有遗漏可加入的木头,因此总的操作数是最少的。
然后我们就要开始求单调链的数量,这里用到一个定理:(这个定理我不知道啊。。。QAQ)
对于Dilworth定理,给出如下解释:
一种Dilworth定理的等价表述是:在有穷偏序集中,任何反链最大元素数目等于任何将集合到链的划分中链的最小数目。一个关于无限偏序集的理论指出,在此种情况下,一个偏序集具有有限的宽度w,当且仅当它可以划分为最少w条链。
可知,对于已经排序好的数据,上升单调链条数=下降单调链的最大长度(任何反链最大元素数目等于任何将集合到链的划分中链的最小数目)
所以这道题最终转化成了求最长下降子序列的问题
代码如下:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<iostream> 2 #include<cstdio> 3 #include<algorithm> 4 using namespace std; 5 struct node 6 { 7 int l,w; 8 }wood[10000]; 9 bool cmp(node a,node b) 10 { 11 if(a.l==b.l) return a.w<b.w; 12 return a.l<b.l; 13 } 14 int dp[10000]; 15 int T,n; 16 void solve() 17 { 18 for(int i=1;i<=n;i++) 19 { 20 dp[i]=1; 21 for(int j=1;j<=i;j++) 22 { 23 if(wood[j].w>wood[i].w) 24 dp[i]=max(dp[i],dp[j]+1); 25 } 26 //printf("now:%d ",dp[i]); 27 } 28 } 29 int main() 30 { 31 //freopen("in.txt","r",stdin); 32 cin>>T; 33 for(int i=1;i<=T;i++) 34 { 35 cin>>n; 36 for(int j=1;j<=n;j++) 37 { 38 scanf("%d %d",&wood[j].l,&wood[j].w); 39 } 40 sort(wood+1,wood+1+n,cmp); 41 //for(int j=1;j<=n;j++) printf("%d %d ",wood[j].l,wood[j].w); 42 solve();int ans=0; 43 for(int j=1;j<=n;j++) ans=max(ans,dp[j]); 44 printf("%d ",ans); 45 } 46 return 0; 47 }