心急的C小加（两种解法）

心急的C小加

时间限制：1000 ms  |  内存限制：65535 KB

难度：4

 

描述

C小加有一些木棒，它们的长度和质量都已经知道，需要一个机器处理这些木棒，机器开启的时候需要耗费一个单位的时间，如果第i+1个木棒的重量和长度都大 于等于第i个处理的木棒，那么将不会耗费时间，否则需要消耗一个单位的时间。因为急着去约会，C小加想在最短的时间内把木棒处理完，你能告诉他应该怎样做 吗？

输入

第一行是一个整数T(1<T<1500)，表示输入数据一共有T组。
每组测试数据的第一行是一个整数N（1<=N<=5000）,表示有N个木棒。接下来的一行分别输入N个木棒的L，W（0 < L ,W <= 10000），用一个空格隔开，分别表示木棒的长度和质量。

输出

处理这些木棒的最短时间。

样例输入

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

样例输出

2
1
3

先贴下自己的dp超时代码：

 #include<stdio.h>
 #include<algorithm>
 #define MAX(x,y) x>y?x:y
 using namespace std;
 struct Case{
     int L,W;
 }strick[];
 int cmp(Case a,Case b){
     if(a.L==b.L)return a.W<b.W;
     else return a.L<b.L;
 }
 int main(){
     int T,N,dp[],max;
     scanf("%d",&T);
     while(T--){
         scanf("%d",&N);
         for(int i=;i<N;++i){dp[i]=;
         scanf("%d%d",&strick[i].L,&strick[i].W);
         }
         sort(strick,strick+N,cmp);
         max=;
         for(int i=;i<N;++i){
             for(int j=;j<=i;++j){
                 if(strick[j].W>strick[i].W)dp[i]=dp[j]+;
                 max=MAX(dp[i],max);
             }
         }
         printf("%d\n",max);
     }
     return ;
 }

借助大神理解解写的代码：

 #include<stdio.h>

 #include<algorithm>

 using namespace std;

 struct Case{

     int L,W,dis;

 };

 struct Case stick[];

 int cmp(Case a,Case b){

     if(a.L==b.L)return a.W<b.W;

     else return a.L<b.L;

 }

 int main(){int T,N,t,flot;

 scanf("%d",&T);

 while(T--){

     scanf("%d",&N);

     for(int i=;i<N;++i)scanf("%d%d",&stick[i].L,&stick[i].W),stick[i].dis=;

     sort(stick,stick+N,cmp);flot=;

     for(int i=;i<N;++i){

         if(stick[i].dis==)continue;

         t=stick[i].W;

         for(int j=i+;j<N;++j){

             if(stick[j].dis&&t<=stick[j].W)stick[j].dis=,t=stick[j].W;

         }

         flot++;

     }

     printf("%d\n",flot);

 }

     return ;

 }

/*题解：


结构体排序，一级排木棒的重量，二级排木棒的长度，均由小到大。


进行了多少次外层循环，就是花费的分钟数。


*/ 

若排序后，如图：

则，花费了2分钟。 

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14880    Accepted Submission(s): 6105

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

2 1 3

 两题一样.....