HDU 题目1051 Wooden Sticks

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11126    Accepted Submission(s): 4579

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

 

此题意思即使按照题意，一开始题意没搞清楚，理解为小的木棍不用时间，实际上是在原来安装那的木棍上安装不用花时间。。。

#include <stdio.h>struct stick{    int len;    int weight;    int cnt;}woolen[5005],t;int main(void){  // freopen("F.txt","r",stdin);    int i,j,n,m,k,sum,count;    while(scanf("%d",&n)!=EOF)    {        for(i=1;i<=n;i++)        {            scanf("%d",&m);            for(j=1;j<=m;j++)            {                scanf("%d%d",&woolen[j].len,&woolen[j].weight);                woolen[j].cnt=0;            }          //  for(j=1;j<=m;j++)            //  printf("%d %d\n",woolen[j].len,woolen[j].weight);            for(j=1;j<=m;j++)            {                for(k=1;k<=m-j;k++)                {                    if(woolen[k].len>woolen[k+1].len)                    {                        t=woolen[k];                        woolen[k]=woolen[k+1];                        woolen[k+1]=t;                    }                    if(woolen[k].len==woolen[k+1].len&&woolen[k].weight>woolen[k+1].weight)                    {                         t=woolen[k];                        woolen[k]=woolen[k+1];                        woolen[k+1]=t;                    }                }            }         //  for(j=1;j<=m;j++)             // printf("%d %d\n",woolen[j].len,woolen[j].weight);          /*  for(j=1;j<=m;j++)            {                if(woolen[j].cnt==0)                {                   for(k=1;k<j;k++)                {                    if(woolen[j].weight>=woolen[k].weight)                       {                            woolen[j].cnt=1;                            break;                       }                }                }            }*/            for(j=1;j<=m;j++)            {                if(woolen[j].cnt==0)                {                    t=woolen[j];                }                for(k=j+1;k<=m;k++)                {                    if(woolen[k].cnt==0)                    {                        if(t.weight<=woolen[k].weight)                        {                            t=woolen[k];                            woolen[k].cnt=1;                        }                    }                }            }            sum=0;            for(j=1;j<=m;j++)                {                    if(woolen[j].cnt==0)                sum++;                }                printf("%d\n",sum);        }    }    return 0;}