HDU 1051 Wooden Sticks(貪心演算法)

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9018    Accepted Submission(s): 3680


Problem DescriptionThere 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). 

InputThe 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. 

OutputThe 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

 

SourceAsia 2001, Taejon (South Korea) 這個題目的貪心演算法比較簡單，首先對L進行排序，那麼這樣後面的一定在前面的後處理最小的按照貪心來說第一個處理，處理完成之後向後掃描把這一次能處理的全部處理掉不能處理的留下來下一次處理最小的時候仔處理，這樣一直到所有全部處理完成結束！

#include <iostream>#include <string.h>#include <stdio.h>#include <algorithm>using namespace std;struct point{    int l;    int w;}po[5500],temp;int cmp(const void *a,const void *b){  if((*(point*)a).l == (*(point*)b).l)  return (*(point*)a).w < (*(point*)b).w ? -1:1;  return (*(point*)a).l < (*(point*)b).l ? -1:1;}int main(){    int t;    int n,i,j;    int ans;    int pos,k,real_n;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(i=0;i<n;i++)        scanf("%d%d",&po[i].l,&po[i].w);        qsort(po,n,sizeof(po[0]),cmp);        ans=0;        k=0;        while(n>0)        {            ans++;            k=0;            temp=po[0];            for(i=1;i<n;i++)            if(po[i].w>=temp.w)            temp=po[i];            else            po[k++]=po[i];            n=k;        }        printf("%d\n",ans);       // for(i=0;i<n;i++)       // printf("%d %d\n",po[i].l,po[i].w);    }    return 0;}