HDU Wooden Sticks (貪心)

Wooden Sticks

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 19 Accepted Submission(s) : 7

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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) AC代碼：

#include 
#include 
#include 
#include 
using namespace std;
struct wooden{
    int l,w;
};
wooden my[5010];
bool comp(wooden a,wooden b){
    if(a.l>b.l)return 1;
    else if(a.l==b.l)
        return a.w>b.w;
    else return 0;
}
int main(){
    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        int i=0,j;
        while(i>my[i++].l>>my[i].w;
        sort(my,my+n,comp);
        //for(i=0;i=my[i].l&&my[j].w>=my[i].w){
            out--;
            my[j].l=my[i].l;
            my[j].w=my[i].w;
            my[i].l=0;
            my[i].w=0;
            break;
            }
        }
        // for(i=0;i