HUD1051 解题报告

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22263    Accepted Submission(s): 8959

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

213

 

Source

Asia 2001, Taejon (South Korea)

 

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解法：贪心

将所有的木棒按照长短排列，长的在前，如果长度相同，则重的在前面，然后从第一根开始取

接下来的一根的长度和质量都不能超过前面的那一根，一直取到没有满足条件的木棒为止

然后从第一根没有取过的木棒开始，依次从没有取过的木棒中挑选符合要求的

#include<iostream>#include<algorithm>using namespace std;struct wood{    int l;    int w;    bool used;};const int maxn=5000+50;const int maxlw=10000;wood woods[maxn];int n;bool cmp(const wood &w1,const wood &w2);void solve();int main(){    int t;    cin>>t;    for(int j=0;j<t;j++)    {        cin>>n;        for(int i=0;i<n;i++)        {            cin>>woods[i].l>>woods[i].w;            woods[i].used=false;        }        solve();    }}bool cmp(const wood &w1,const wood &w2){    if(w1.l!=w2.l) return w1.l>w2.l;    return w1.w>w2.w;}void solve(){    sort(woods,woods+n,cmp);    int ans=0;    int res=n; //剩下的数目    int save_fi=-1;    while(res)    {        ans++;        int fi;        for(fi=save_fi+1;fi<n;fi++) //找到第一个没有用过的棒子            if(!woods[fi].used)                break;        woods[fi].used=true;        save_fi=fi;        res--;        wood cmpwood=woods[fi];        for(int i=fi+1;i<n;i++)        {            if(!woods[i].used&&woods[i].w<=cmpwood.w)            {                woods[i].used=true;                cmpwood=woods[i];                res--;            }        }    }    cout<<ans<<endl;}