贪心之Wooden Sticks

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 
// 贪心算法---先排序---后选择第一个没有用过的木头一次向后找，用掉所有可以用掉的木头，然后返回第一个没用过的木头继续找#include<iostream>#include<string.h>#include<stdio.h>#include<ctype.h>#include<algorithm>#include<stack>#include<queue>#include<set>#include<math.h>#include<vector>#include<deque>#include<list>using namespace std;struct mg{    int l;    int z;}p[5050];bool cmp(mg a,mg b){    if(a.l==b.l)    return a.z<b.z;    else    return a.l<b.l;}int s[5050];int main(){    int T,n,i,j;    int count=0;    bool flag;    scanf("%d",&T);    while(T--)    {        count=0;        scanf("%d",&n);        for(i=0;i<n;i++)        cin>>p[i].l>>p[i].z;        sort(p,p+n,cmp);        memset(s,0,sizeof(s));        s[0]=1;        int jl=0;//用来记录第一次没用的木块        count=0;        while(jl<n)        {            count++;            for(i=jl+1,j=jl,flag=1;i<n;i++)            {                if(s[i])//标记为1的就跳过                continue;                if(p[i].l>=p[j].l&&p[i].z>=p[j].z)//两个条件都满足，才能分为一组                {                    j=i;                    s[i]=1;                }                else                {                    if(flag)                    {                        jl=i;                        flag=0;                    }                }            }            if(flag)//如果走了一遍都符合就，结束            break;        }        printf("%d\n",count);    }    return 0;}