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所谓的偏序定理 说的就是 Dilworth定理
给定n个二元组(x, y)，问存在最少多少个划分使得每个划分里面的二元组都满足x1 <= x2并且y1 <= y2。
如果定义x1 <= x2 && y1 <= y2为偏序关系的话，那么问题就转化成求这个集合的链的最少划分数。可以通过找最长反链长度来解决，这里的反链关系是x1 > x2 || y1 > y2。如果把n个二元组按照x递增排序，相同的x按照y递增排序，那么我们只需对y找到一个最长递减子序列就是所求的答案，复杂度O(nlogn)。对于相同的x之所以按照y递增排序是因为这里偏序关系带等号，这样相同的x其实可以划分到一起，把y按照递增排序就可以使得相同的x最多只选择一个y。
还有的题目要求满足x1 < x2 && y1 < y2，这就需要把偏序关系相应修改。修改之后对于相同的x，每一个都会被划分到不同的集合（因为相等是不满足偏序关系的），所以这里的排序关系要改一下，x相同的y要按照降序排列，这样求一个最长不递增子序列就是答案，y递减保证可能会有多个x相同的二元组选入到结果中。
poj 1065
| Language: Wooden Sticks
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 ) . 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 |
题意不再赘述
木棍的重量和长度只要满足>= 就可以了 所以我们定义排序规则 重量和长度都是按升序排列
#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string.h>
#include <string>
#include <vector>
#include <queue>#define MEM(a,x) memset(a,x,sizeof a)
#define eps 1e-8
#define MOD 10009
#define MAXN 5010
#define MAXM 100010
#define INF 99999999
#define ll __int64
#define bug cout<<"here"<<endl
#define fread freopen("ceshi.txt","r",stdin)
#define fwrite freopen("out.txt","w",stdout)using namespace std;int Read()
{char c = getchar();while (c < '0' | c > '9') c = getchar();int x = 0;while (c >= '0' && c <= '9') {x = x * 10 + c - '0';c = getchar();}return x;
}void Print(int a)
{if(a>9)Print(a/10);putchar(a%10+'0');
}struct node
{int l,w;void input(){scanf("%d%d",&l,&w);}bool operator<(const node& p)const{if(l==p.l) return w<p.w;else return l<p.l;}
}no[MAXN];int vis[MAXN];int main()
{
// fread;int tc;scanf("%d",&tc);int n;while(tc--){scanf("%d",&n);for(int i=0;i<n;i++)no[i].input();sort(no,no+n);
// for(int i=0;i<n;i++)
// cout<<no[i].l<<" "<<no[i].w<<endl;MEM(vis,0);int ans=0;for(int i=0;i<n;i++){if(vis[i])continue;vis[i]=1;node p;p.l=no[i].l; p.w=no[i].w;for(int j=i+1;j<n;j++){if(vis[j]) continue;if(no[j].l>=p.l&&no[j].w>=p.w){vis[j]=1;p.l=no[j].l; p.w=no[j].w;}}ans++;}printf("%d\n",ans);}return 0;
}
poj 3636
| Language: Nested Dolls
Description Dilworth is the world's most prominent collector of Russian nested dolls: he literally has thousands of them! You know, the wooden hollow dolls of different sizes of which the smallest doll is contained in the second smallest, and this doll is in turn contained in the next one and so forth. One day he wonders if there is another way of nesting them so he will end up with fewer nested dolls? After all, that would make his collection even more magnificent! He unpacks each nested doll and measures the width and height of each contained doll. A doll with width w1 and height h1 will fit in another doll of width w2 and heighth= if and only if w1 < w2 and h1 < h2. Can you help him calculate the smallest number of nested dolls possible to assemble from his massive list of measurements? Input On the first line of input is a single positive integer 1 ≤ t ≤ 20 specifying the number of test cases to follow. Each test case begins with a positive integer 1 ≤ m ≤ 20000 on a line of itself telling the number of dolls in the test case. Next follow 2m positive integers w1, h1,w2, h2, ... ,wm, hm, where wi is the width and hi is the height of doll number i. 1 ≤ wi, hi ≤ 10000 for all i. Output For each test case there should be one line of output containing the minimum number of nested dolls possible. Sample Input 4 3 20 30 40 50 30 40 4 20 30 10 10 30 20 40 50 3 10 30 20 20 30 10 4 10 10 20 30 40 50 39 51 Sample Output 1 2 3 2 |
这题题意与上题类似 最大不同就是 宽度和高度必须要是大于才是满足的
所以我们定义排序规则 宽度按升序 高度按降序 这样才会保证在宽度相同时 不会出现覆盖的情况
另外 这道题的数据比较大 我们可以记录每一个套娃最外面的宽度和高度
如果无法把枚举的套娃覆盖在之前的套娃上 那么就要多放一个。。。
#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string.h>
#include <string>
#include <vector>
#include <queue>#define MEM(a,x) memset(a,x,sizeof a)
#define eps 1e-8
#define MOD 10009
#define MAXN 20010
#define MAXM 100010
#define INF 99999999
#define ll __int64
#define bug cout<<"here"<<endl
#define fread freopen("ceshi.txt","r",stdin)
#define fwrite freopen("out.txt","w",stdout)using namespace std;int Read()
{char c = getchar();while (c < '0' | c > '9') c = getchar();int x = 0;while (c >= '0' && c <= '9') {x = x * 10 + c - '0';c = getchar();}return x;
}void Print(int a)
{if(a>9)Print(a/10);putchar(a%10+'0');
}struct node
{int w,h;void input(){scanf("%d%d",&w,&h);}bool operator<(const node &p)const{if(w==p.w) return h>p.h;else return w<p.w;}
}no[MAXN];
node e[MAXN];int main()
{
// fread;int tc;scanf("%d",&tc);while(tc--){int n;scanf("%d",&n);for(int i=0;i<n;i++)no[i].input();sort(no,no+n);int ans=0;for(int i=0;i<n;i++){int flag=0;for(int j=0;j<ans;j++){if(no[i].w>e[j].w&&no[i].h>e[j].h){flag=1;e[j]=no[i];break;}}if(!flag)e[ans++]=no[i];}printf("%d\n",ans);}return 0;
}
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