HDU1051:Wooden Sticks

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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

21
3
题意是：将木棍放在机器里处理，第一根需要一分钟下一根的长度和重量如果大于等于前边放入的长度和重量，就不用费时间，否则需要一分钟调试，计算给出一组数的最少时间！   
先用sort从小到大排序  算出非严格递增子序列的个数！  
换个思路也就是求最大递减子序列的长度  这里不过多解释  多思考 举几个例子就明白了
#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int dp[10001];struct node{int x,y;}s[10001];bool cmp(node a,node b){if(a.x !=b.x )return a.x <b.x ;return a.y <b.y ;}int main(){    int n,t;    scanf("%d",&t);    while(t--)    {    scanf("%d",&n);    for(int i=0;i<n;i++)    {    scanf("%d%d",&s[i].x ,&s[i].y );    dp[i]=1;}sort(s,s+n,cmp);for(int i=0;i<n;i++){for(int j=0;j<i;j++){if(s[j].y >s[i].y &&dp[i]<dp[j]+1){dp[i]=dp[j]+1;}}}int max=0;for(int i=0;i<n;i++){if(dp[i]>max){max=dp[i];}}printf("%d\n",max);}return 0;}