Hdu1051 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

2

1

3

题意为一台机器处理木棒，
1.处理第一根木棒需要1分钟；
2.若之后处理的木棒长度以及重量均大于等于之前的木棒，则不需要花费时间；
求最少耗时。
很显然是一道贪心题，
我们首先对木棒进行排序，用u[i]来表示木棒是否被处理；
那么取第一根木棒进行处理，对之后满足条件2的木棒均进行处理，
这样在回过头去找未被处理的第一根木棒继续进行上述操作。

#include <cstdio>#include <iostream>#include <cmath>#include <cstring>#include <algorithm>using namespace std;struct mubang{    int l;    int w;}a[5005];bool cmp(mubang k1,mubang k2){    if(k1.l==k2.l)        return k1.w<k2.w;    else        return k1.l<k2.l;}int main(){    int t,n,i,s,ans,ll,ww;    int u[5005];    scanf("%d",&t);    while(t--){        memset(u,0,sizeof(u));        scanf("%d",&n);        for(i=0;i<n;i++){            scanf("%d%d",&a[i].l,&a[i].w);        }        sort(a,a+n,cmp);        s=0;        ans=0;        while(s<n){            if(u[s]==0){                ll=a[s].l;                ww=a[s].w;                ans++;        for(i=s;i<n;i++){            if(u[i]==0&&a[i].l>=ll&&a[i].w>=ww){                ll=a[i].l;                ww=a[i].w;                u[i]=1;            }        }            }            else s++;        }        printf("%d\n",ans);    }    return 0;}