Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2471 Accepted Submission(s): 955 

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)

/*贪心算法*/#include<stdio.h>#include<stdlib.h>#include<string.h>struct wooden{   int length;   int weight;   int flag;};int cmp(const void *a,const void *b){    struct wooden *c=(struct wooden *)a;//类型转换    struct wooden *d=(struct wooden *)b;    if(c->length==d->length)    {        return c->weight - d->weight;//按照weight大小排序 由小到大    }    else return c->length - d->length;//由小到大}int main(){    int num=0;    int n=0;    int i=0,j=0,k=0;    int max=0;    int time=0;    struct wooden wood[5001];    scanf("%d",&num);    for(i=0;i<num;i++)    {       memset(wood,0,sizeof(struct wooden)*5001);//标志位清0       scanf("%d",&n);       for(j=0;j<n;j++)       {           scanf("%d%d",&wood[j].length,&wood[j].weight);       }        qsort(wood,n,sizeof(struct wooden),cmp);//全部按照从小到大排序        time=0;        for(j=0;j<n;j++)        {           if(wood[j].flag) continue;//等于1表示处理过了           max=wood[j].weight;           for(k=j+1;k<n;k++)           {               if(wood[k].weight>=max && !wood[k].flag )               {                   max=wood[k].weight;                   wood[k].flag=1;               }           }           time++;         }       printf("%d\n",time);    }    return 0;}