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最新推荐文章于 2018-11-18 19:34:23 发布

Wiking__acm

最新推荐文章于 2018-11-18 19:34:23 发布

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本文链接：https://blog.csdn.net/Wiking__acm/article/details/12881949

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本博客讨论了如何从给定的俄罗斯套娃娃尺寸列表中，计算出使用最少数量的套娃组合来达到相同的嵌套效果。通过排序和求最长不减序列，解决了这一问题。

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Nested Dolls

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2153 Accepted Submission(s): 631

Problem 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 height h2 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

Source

2008 “Insigma International Cup” Zhejiang Collegiate Programming Contest - Warm Up（3）

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lcy

题意就是把原来的序列划分成最少的集合，集合内的元素满足偏序关系。开始想最大匹配，然后发现时间不够。其实这是一个二维的“导弹拦截系统”，并且这里必须是严格的小于关系。做法是先把原来的元素按宽排序，对于宽相等的，必须按长从大到小排序，然后得到一个新的由长组成的序列。对这个序列，求反向最长不减序列的长度就是答案。原因是排在后面的长对应的宽一定大于等于前面的，而等于的情况，为了保证不会误选，使长从大到小排序。再根据最小集合划分等于最长反链长度，就可以求得答案了。不减序列的求法，就是把原来二分中的lower_bound改成upper_bound即可。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <string>
#include <algorithm>
#include <map>
#include <vector>
#include<queue>
#include<set>
#include<cmath>
using namespace std;
typedef long long LL;
typedef pair<double,int> P;
const int maxn = 20000 + 5;
const int INF = 1000000000;

struct Node{
    int x,y;
}p[maxn];

bool cmp(Node a,Node b){
    if(a.x == b.x) return a.y > b.y;
    return a.x < b.x;
}

int a[maxn];
int dp[maxn];
int n;

void solve(){
    fill(dp,dp+n,INF);
    for(int i = 0;i < n;i++){
        *upper_bound(dp,dp+n,a[i]) = a[i];
    }
    printf("%d\n",lower_bound(dp,dp+n,INF)-dp);
}

int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        for(int i = 0;i < n;i++){
            int x,y;
            scanf("%d%d",&p[i].x,&p[i].y);
        }
        sort(p,p+n,cmp);
        for(int i = 0;i < n;i++) a[i] = p[n-i-1].y;
        solve();
    }
    return 0;
}