HDU 1051.Wooden Sticks【贪心】【8月26】

Wooden Sticks

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

有些木头长度len，重量wei已知。如果前一根的长度>=下一根的长度并且前一跟的重量>=下一根的重量，则处理下一根木头不花费时间。求最少花费的时间。贪心，排序之后遍历。代码如下：

#include<cstdio>#include<algorithm>using namespace std;struct ss{    int len,wei;    bool fla;};bool cmp(ss x,ss y){    if(x.len<y.len) return true;    else if(x.len==y.len&&x.wei<y.wei) return true;    else return false;}int main(){    int T,n;    scanf("%d",&T);    while(T--){        ss f[5010];        int sum=0;        scanf("%d",&n);        for(int i=0;i<n;i++){            scanf("%d %d",&f[i].len,&f[i].wei);            f[i].fla=false;        }        sort(f,f+n,cmp);        for(int i=0;i<n;i++)        if(!f[i].fla){            f[i].fla=true;            sum++;            int wei=f[i].wei;            for(int j=i+1;j<n;j++)            if(!f[j].fla&&f[j].wei>=wei){                f[j].fla=true;                wei=f[j].wei;            }        }        printf("%d\n",sum);    }    return 0;}