http://acm.hdu.edu.cn/showproblem.php?pid=1051

Wooden Sticks（护眼模式）

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 18540    Accepted Submission(s): 7556 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   Source Asia 2001, Taejon (South Korea)  

题意：

给定一堆需要加工的乱序的木头，目前加工的这一根的长度和重量比之前加工的那一根都大，就不需要升级机器，

升级机器花费的时间是一分钟。问最后至少需要花费多少分钟。

思路： LIS 和 dp 的结合，很基础的题了，今天第二天学习这个算法（其实也不算新的学习），感觉还是需要慢慢的去自己进行演示，理解。

Code：

#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; const int MYDD=1103+5e3; struct Q { int length;//长度 int weight;//密度 } wooden[MYDD]; int dp[MYDD];//当前木头之前的时间数 bool cmp(Q x,Q y) { if(x.length!=y.length) return x.length<y.length; return x.weight<y.weight; } int main() { int TT; scanf("%d",&TT); while(TT--) { int n,i,j; memset(dp,0,sizeof(dp));//dp[]的初始化 scanf("%d",&n); for(j=1; j<=n; j++) scanf("%d%d",&wooden[j].length,&wooden[j].weight); sort(wooden+1,wooden+n+1,cmp); for(j=1; j<=n; j++) for(i=1; i<=j; i++) if(wooden[j].weight<wooden[i].weight) if(dp[j]<dp[i]+1) dp[j]=dp[i]+1; sort(dp,dp+MYDD);//寻找最大值 printf("%d\n",dp[MYDD-1]+1);//dp[]初始化是 0 } return 0; }