hdu1051 Wooden Sticks（LIS动态规划or贪心）

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 18663    Accepted Submission(s): 7640

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

 

题意是：将木棍放在机器里处理，第一根需要一分钟，剩余的如果大于等于前边放入的长度和重量，就不用费时间，否则需要一分钟，计算给出一组数的最少时间！

一.dp

先对木棍进行排序，先按长度从小到大，长度相同，再按重量从小到大，排序之后，可以知道第i根木棍总不比第i-1根短，只是就只需要考虑重量就行，如果j根木棍比j-1根长，就可得需要另开多花费一分钟，  dp[i]=dp[j]+1就行，最后求的1-n中dp[i]的最大值

#include<stdio.h>  #include<string.h>  #include<algorithm>  using namespace std;  struct st  {      int l,w;  }a[5010];  int dp[5010];  int cmp(st x,st y)  {      if(x.l!=y.l)      return x.l<y.l;//按照筷子的长度递增排序，判断筷子的重量是否是递增的，若递增满足题意，否则+1；       else       return x.w<y.w;  }  int main()  {      int t,n;      scanf("%d",&t);      while(t--)      {          scanf("%d",&n);          for(int i=0;i<n;i++)          {              scanf("%d%d",&a[i].l,&a[i].w);              dp[i]=1;          }          sort(a,a+n,cmp);          for(int i=1;i<n;i++)          {              for(int j=0;j<i;j++)              {                  if(a[j].w>a[i].w&&dp[i]<dp[j]+1)//若满足dp[i]<dp[j]+1表示i和j在相同的序列中                                              // 若在相同的序列中dp[i]就要+1;                  {                                           dp[i]=dp[j]+1;                  }              }          }          int maxn=0;          for(int i=0;i<n;i++)          {              if(dp[i]>maxn)                 maxn=dp[i];           }          printf("%d\n",maxn);      }      return 0;  }  

二 贪心

基本和dp差不多，也是用序列来完成，排序之后就只看重量了，如果后者比前者大，更新较大数，反之再开一个新序列

#include<stdio.h>  #include<string.h>  #include<algorithm>  using namespace std; struct st{int l,w;}a[5005];int s[5005];   bool cmp(st a,st b)//排序 {if(a.l!=b.l)     return a.l<b.l;else return a.w<b.w;}int main(){int t;scanf("%d",&t);while(t--){int n;scanf("%d",&n);for(int i=0;i<n;i++){scanf("%d%d",&a[i].l,&a[i].w);s[i]=1;}sort(a,a+n,cmp);int exist=1;  s[0]=a[0].w;for(int i=1;i<n;i++){int flag=1;for(int j=0;j<exist;j++){if(s[j]<=a[i].w){flag=0;s[j]=a[i].w;break;}        }if(flag)  //前面所有重量大于当前重量，再开一个序列  s[exist++]=a[i].w;}printf("%d\n",exist);}return 0;}