HDU 1051 Wooden Sticks(贪心算法)

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 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)

 

这个题目的贪心算法比较简单，首先对L进行排序，那么这样后面的一定在前面的后处理

最小的按照贪心来说第一个处理，处理完成之后向后扫描把这一次能处理的全部处理掉

不能处理的留下来下一次处理最小的时候仔处理，这样一直到所有全部处理完成结束！

#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;}