C - 贪心3 HDU - 1051 题解

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
问题描述
题目给出n个木棍的长度和重量，当机器处理完长度为L，重量为W的木棍后再处理长度为L1（L<=L1),重量为W1（W<=W1)的木棍就不需要处理时间，否则就需要一定的处理时间，根据给出的数据求出最短的处理时间。
伪代码

class Node{//代表一根木棍 
木棍的长：L 
木棍的重量：W 
自定义<:Ltrue||L==L1&&Wtrue否则放回false
}
1.将数据存入到vector v中
2.对v排序
3.定义变量vector v_temp用来某一处理序列的最大值
4.将v中的首元素插入到v_temp中
把v中从第二个开始的所有元素和v_temp中的所有元素按顺序从小到达比较（即一个二重循环）如果当前v中的元素比v_temp中的所有元素都要小，则插入成为v_temp中的元素，否则将第一个比v中元素小的v_temp中的元素跟新成为当前的v中的元素。
5.最后v_temp的大小即为最后的结果。

代码

#include
#include
#include
#include
using namespace std;
class Node{
public:
    int l,w;
    Node(int l1,int w1):l(l1),w(w1){}
    bool operator<(const Node &N){
        if(lreturn true;
        else if(l==N.l&&wreturn true;
        else return false;
    }
};
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        int n;
        scanf("%d",&n);
        vector v;
        while(n--){
            int l,w;
            scanf("%d %d",&l,&w);
            v.push_back(Node(l,w));
        }
        sort(v.begin(),v.end());
        vector v_temp;
        v_temp.push_back(v[0]);
        for(int i=1;ifor(int j=0;jif(v[i].l>=v_temp[j].l&&v[i].w>=v_temp[j].w){
                    v_temp[j].l=v[i].l;
                    v_temp[j].w=v[i].w;
                    break;
                }
                else {
                    j++;
                    if(j==v_temp.size())
                    v_temp.push_back(v[i]);
                }
            }
            cout<return 0;
}