YT04-贪心课堂练习-1005—Wooden Sticks-（6.14日-烟台大学ACM预备队解题报告）

Wooden Sticks

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 19   Accepted Submission(s) : 7

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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)

计145 王忠

•I）Sort函数有三个参数：

•（1）第一个是要排序的数组的起始地址。

•（2）第二个是结束的地址（最后一位要排序的地址）

•（3）第三个参数是排序的方法，可以是从大到小也可是从小到大，还可以不写第三个参数，此时默认的排序方法是从小到大排序

#include <iostream>#include <stdio.h>#include <algorithm>using namespace std;struct node{    int l,w;} a[10001],flag[10001];bool cmp(node x,node y){    return x.l!=y.l ? x.l<y.l : x.w<y.w ;}int main(){    int t,n,m,i,j;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(i=0; i<n; i++)        {            scanf("%d %d",&a[i].l,&a[i].w);        }        sort(a,a+n,cmp);   //按l升序排        m=0;        flag[m]=a[0];   //结构体既然可以直接=        for(i=1; i<n; i++)        {            for(j=0; j<=m; j++) //在flag中w小的总是在后面，所以不用再根据w排序            {                if(flag[j].w<=a[i].w)   //更新原有数据                {                    flag[j]=a[i];                    break;                }            }            if(j>m)   //加入新数据            {                flag[++m]=a[i];            }        }        printf("%d\n",m+1);    }    return 0;}