杭电1051 Wooden Sticks (贪心)

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Wooden Sticks

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 8   Accepted Submission(s) : 4

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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
/*心得:wa了好几次,看了包子的代码才知道自己思路错了注意:不能直接按 l 或 w 排序后那后一个和前一个比较*/#include<stdio.h>#include<string.h>#include<algorithm>#define MAX 5000+10using namespace std;struct stick{     int w,l;  }s[MAX];bool cmp(stick a,stick b){     if(a.l==b.l&&a.w>=b.w)return 1;     else if(a.l>b.l)return 1;     return 0;}int main(){    int T;    scanf("%d",&T);    while(T--)    {              int N;     scanf("%d",&N);     for(int i=0;i<N;i++)     scanf("%d%d",&s[i].l,&s[i].w);     int t=0;      sort(s,s+N,cmp);     bool vis[MAX];     memset(vis,0,sizeof(vis));     for(int i=0;i<N;i++)     {         if(vis[i])continue;          else t++;                       int ww=s[i].w;          for(int j=i+1;j<N;j++)         {             if(ww>=s[j].w&&!vis[j])//此处没有vis[]会重复访问             {                ww=s[j].w;                vis[j]=1;                            }             }         }         printf("%d\n",t);    }return 0;}

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