HDU1051切割木头

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16938    Accepted Submission(s): 6910

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
//思路：题目大概就是讲一个木头放入一个机器切割需要1分钟，如果后面放入的木头高度宽度均小于前者，则不要时间，否则一分钟#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <set>using namespace std;struct node{    int x,y,flag;};int main(){    int n,t;    struct node w[10001],temp;    while(scanf("%d",&t)!=EOF)    {        while(t--)        {            scanf("%d",&n);            for(int i=0; i<n; i++)            {                scanf("%d%d",&w[i].x,&w[i].y);                w[i].flag=1;            }            for(int i=0;i<n;i++)            {                for(int j=i+1;j<=n;j++)                {                    if(w[i].x>w[j].x)//x从小到大排列                    {                        temp=w[i];                        w[i]=w[j];                        w[j]=temp;                    }                    else if(w[i].x==w[j].y&&w[i].x>w[j].y)//如果x相同，看y排                    {                        temp=w[i];                        w[i]=w[j];                        w[j]=temp;                    }                }            }            int time=0;            for(int i=0;i<n;i++)            {                if(w[i].flag)                {                    temp=w[i];                    time++;                    for(int j=i+1;j<=n;j++)                    {                        if(w[j].flag)                        if(temp.y<=w[j].y)//直接拍y就ok了，因为x上面排好了                        {                            temp=w[j];                            w[j].flag=0;//记得标记使用过了                        }                    }                }            }            printf("%d\n",time);        }    }    return 0;}