HDU 1051 Wooden Sticks

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21459    Accepted Submission(s): 8656

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

解题思路：只有wi和li都大于前者才不用消耗时间来准备。所以排序一下，循环多次扫，每次扫能完成的，统计扫描的次数。

代码如下：

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;struct Node{int l,w;}data[5005];int vi[5005];int n;int cmp(Node a,Node b){if(a.l!=b.l)return a.l<b.l;elsereturn a.w<b.w;}int main(){int t;int ans;scanf("%d",&t);while(t--){ans=0;memset(vi,0,sizeof(vi));scanf("%d",&n);for(int i=0;i<n;i++)scanf("%d%d",&data[i].l,&data[i].w);sort(data,data+n,cmp);for(int i=0;i<n;i++){if(!vi[i]){vi[i]=1;int l=data[i].l; int w=data[i].w;for(int j=i+1;j<n;j++){if(!vi[j] && data[j].l>=l && data[j].w>=w){vi[j]=1;l=data[j].l;//注意比较对象的改变。 w=data[j].w;}}ans++;}}printf("%d\n",ans);}return 0;}