Wooden Sticks HDU

 Wooden Sticks

 HDU -1051 

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 ti
me 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

AC代码：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;struct note{    int u,v;}e[10000];int cmp(note A,note B){    if(A.u!=B.u)//在长度不等时，按长度从大到小排        return A.u>B.u;    else        return A.v>B.v;//长度相等时，按重量从大到小排}int n,sum,flag;int book[10000];int main(){    int T;    scanf("%d",&T);    while(T--)    {        int i,j;        memset(book,0,sizeof(book));        scanf("%d",&n);        for(i=0;i<n;i++)        {            scanf("%d%d",&e[i].u,&e[i].v);        }        sort(e,e+n,cmp);        int s=0,minn;        for(i=0;i<n;i++)//贪心，，按例子说：5.4.3.2.1排好后按重量排，能         {               //从大到小排几组（重量），就是需要几分钟。            if(book[i]==0)            {                minn=e[i].v;            }            else                continue;            for(j=i+1;j<n;j++)            {                if(e[j].v<=minn&&book[j]==0)                {                    book[j]=1;                    minn=e[j].v;                }            }            s++;        }        printf("%d\n",s);    }}