HDU 1051 Wooden Sticks

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
贪心 看懂题目就容易了
#include <iostream>#include <algorithm>using namespace std;struct q{    int w,p,ji;}bao[10000];bool cmp(q a,q b){    return a.w < b.w || a.w==b.w && a.p<b.p;}int main(){    int n,i,j,m;    int sum;    scanf("%d",&m);    while(m--)    {        scanf("%d",&n);        {            for(i=0;i<n;i++)            {                bao[i].ji=1;                scanf("%d %d",&bao[i].w,&bao[i].p);            }                        sum=0;                        sort(bao,bao+n,cmp);                        // for(i=0;i<n;i++)            //    printf("%d %d\n",bao[i].w,bao[i].p);                        int wei,tou;            for(i=0;i<n;i++)            {                if(bao[i].ji==1)                {                    tou=wei=0;                    for(j=i;j<n;j++)                    {                        if(bao[j].ji==1 && tou<=bao[j].p && wei<=bao[j].w)                        {                            bao[j].ji=0;                            tou=bao[j].p;                            wei=bao[j].w;                        }                    }                    sum++;                }            }            printf("%d\n",sum);        }    }    return 33;}