poj 1065 Wooden Sticks

Wooden Sticks

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 23699 Accepted: 10212

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

213

Source

Taejon 2001

题意：n根木材长l_i重w_i，前一根木材大于后一根的话要浪费一分钟准备机器，求最省方案？

解题思路：先将所有木材按照长度从小到大排序，这样的话我们解决问题时就只需要考虑重量就可以了。然后我们开始寻找每一根木材的递增序列，并记录下这根木材的重量，然后找比它大的木材。找的话是从这根木材以后的几根里找，一是防止重复查找，二是这样的话肯定是按照长度递增的。每找到一根就把这根木材标记一下，以防止重复查找，并且保存这根木材的重量，并且更新重量值。循环结束后，答案数加1。

AC代码：

#include<iostream>#include<algorithm>#include<string.h>using namespace std;struct node{    int l;    int w;}s[5010];bool check[5010];int ans;bool cmp(node x,node y){    return x.l<y.l;}int t;int n;int main(){    cin>>t;    while(t--)    {        memset(check,false,sizeof(check));        ans=0;        cin>>n;        for(int i=1;i<=n;i++)        cin>>s[i].l>>s[i].w;        sort(s+1,s+1+n,cmp);        int biao;        for(int i=1;i<=n;i++)        {            biao=s[i].w;            if(!check[i])            {                for(int j=i+1;j<=n;j++)                {                    if(s[j].w>=biao&&!check[j])                    {                        biao=s[j].w;                        check[j]=true;                    }                }                ans++;            }        }        cout<<ans<<endl;    }    return 0;}