hdoj-1051

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Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18610    Accepted Submission(s): 7605


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
 

Source
Asia 2001, Taejon (South Korea)

感觉是贪心算法,先排序,然后不断地更新最大值,就行了,排序按照长度从小到大,长度相同时
按照重量从小到大,找一个数组存放每次出现的不能加入的值,可以加入就更新最大值


#include<cstdio>#include<algorithm>using namespace std;struct node{int l;int w;}arr[6000];int r[6000]; bool cmp(node a,node b){if(a.l==b.l)return a.w<b.w;return a.l<b.l;}int main(){int t;scanf("%d",&t);while(t--){int n;scanf("%d",&n);int i,j;for(i=0;i<n;++i)scanf("%d%d",&arr[i].l,&arr[i].w);sort(arr,arr+n,cmp);int sum=1;r[0]=arr[0].w;for(i=1;i<n;++i){for(j=0;j<sum;++j){if(arr[i].w>=r[j])//更新最大值 {r[j]=arr[i].w;break;}}if(j==sum)//没有找到比它小的 {r[sum]=arr[i].w;++sum;}}printf("%d\n",sum);}return 0;}


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