HDU 1051 Wooden Sticks

一、题目概述

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23470    Accepted Submission(s): 9526

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

 

 

Source

Asia 2001, Taejon (South Korea)

 

二、题目释义

给你一堆具有长度和重量的木头，让你去机器里加工，机器会有一个一分钟的set时间，首次加工需要set一次，当连续放入的木头不满足后者的长度和重量都大于等于前者时，机器需要set一下，让你求出最短的set时间（不用求出序列）

三、思路分析

我们会非常直观的想到从小到大进行排序，但很遗憾，长度和重量我们都必须考虑，我们无法对两个主元进行排序操作，但是要让机器set时间最短，其必然要满足某一个元素如长度是从小到大排序的，这是最优解的必要条件，那我们便可以对长度进行排序，对重量进行贪心。

确定使用贪心策略，最关键的有两步，一步即设计贪心策略，二即证明由贪心得到的局部最优解最优解即为整体最优解

贪心策略：排序之后扫描一遍stick，将满足条件的stick标记为vis[i] = 1，即已经被使用，不满足条件的stick为未使用，（需要注意的是当遇到不满足的stick时，扫描比较应当是比较不符合前一个与不符合后一个）扫描到尾部之后，再从第一个未使用的stick开始，扫描一遍全部未使用的stick重复之前的操作，直到全部stick都被使用，每扫描一次计数+1

证明：贪心策略所得到为每一次尽可能大的子串，每次对子串取尽可能大，所得到即为最少的子串个数

四、AC代码

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

const int N = 5005;
struct stick
{
    int len,wei;
}a[N];

int vis[N];

bool cmp(stick a, stick b)
{
    if(a.len == b.len)
        return a.wei < b.wei;
    else return a.len < b.len;
}

int main()
{
    int T,n; cin >> T;
    int flag;
    while(T--)
    {
        cin >> n;
        memset(vis,0,sizeof(vis));
        for(int i=0; i<n; i++)
            scanf("%d%d",&a[i].len,&a[i].wei);
        sort(a,a+n,cmp);
        for(int i=0; i<n; i++)
        {
            printf("%d %d ",a[i].len,a[i].wei);
        }
        cout << endl;
        int st=0, cnt=0;
        vis[0] = 1;
        while(st<n)
        {
            cnt++; flag = 1;
            for(int j=st,i=st+1; i<n; i++)
            {
                if(vis[i]) continue;
                if(a[j].wei<=a[i].wei)
                {
                    vis[i] = 1;
                    j = i; // 这一步实现，只有符合了条件，比较的stick才会依次向前迭代
                }
                else
                {
                    if(flag) // 这里的flag用来确定只有在这一遍扫描中，只有第一个未使用的会被标记st，以作为下一次的起点
                    {
                        st = i;
                        flag = 0;
                    }
                }
            }
            if(flag) break;
        }
        cout << cnt << endl;
    }
    return 0;
}