HDU 1051 Wooden Sticks（How to use 快排）

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2331 Accepted Submission(s): 908 

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

#include<stdio.h>#include<iostream>#include<string.h>#include<algorithm>       //1.使用qsort（）需 #include<algorithm> OR #include<stdlib.h> using namespace std;struct tim{    int x,y;}a[11111];int tab[50001];int comp(const void *a,const void *b)  //2.定义比较函数{    tim *x=(tim*)a,*y=(tim*)b;    if(x->x!=y->x)    return x->x-y->x;    return x->y-y->y;}//再举个简单的栗子：/*  int compare(const void *a,const void *b){    return *(int*)a - *(int*)b; //if return *(int*b) - *(int*a) ,then the array will be sortted in decreasing order （递减序列）} */int main(){    int i,j,m,n,k,t;    cin>>k;    while(k--)    {        cin>>n;        for(i=1;i<=n;i++)        cin>>a[i].x>>a[i].y;/*3. 实参1.传要排序的数组  +1表示从第一个开始排序 2.n 数组元素个数 3.一个元素的大小 4.比较函数  OK快排的实现请参照 http://blog.csdn.net/u012400327/article/details/22619497 */        qsort(a+1, n, sizeof(tim), comp);        memset(tab, 0, sizeof(tab));        m=j=0;        while(j<n)        {            m++;            i=1;            while(tab[i]==1)            i++;            int temp1=i;            tab[i]=1;            j++;            i++;        //    cout<<temp1<<' '<<i<<' '<<j<<endl;            while(i<=n)            if(tab[i]==0)            {                {                    if(a[i].x>=a[temp1].x&&a[i].y>=a[temp1].y)                    {                        temp1=i;                        tab[i]=1;                        j++;                    }                }                i++;            }            else            i++;        }        cout<<m<<endl;    }}