poj1065(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 ( 9 , 4 ) , ( 2 , 5 ) , ( 1 , 2 ) , ( 5 , 3 ) , and ( 4 , 1 ) , then the minimum setup time should be 2 minutes since there is a sequence of pairs ( 4 , 1 ) , ( 5 , 3 ) , ( 9 , 4 ) , ( 1 , 2 ) , ( 2 , 5 ) .

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 2n 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

有一些木棍，给出每根的长度和重量，依次处理这些木棍，如果当前处理的木棍长度和重量都大于等于前一根木棍，就不需要花时间，否则花费一个单位时间，安排适当的处理顺序，使得总花费最小。解法：贪心思想，首先对木棍排序，按照长度从小到大排序，相同时重量小的在前。用变量flag标记是否已经被处理过。然后依次处理，处理第i根时如果flag=0，那么时间加1，并且遍历之后的木棍，找出最大的上升子序列，也就是全都不需要花费时间的木棍，标记为flag=1即可。

#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <map>
#include <vector>
#include <list>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <functional>
#include <iomanip>
#include <limits>
#include <new>
#include <utility>
#include <iterator>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cmath>
#include <ctime>
using namespace std;
typedef long long LL;
const int INF = 0x3f3f3f3f;
const double PI = acos(-1.0);
const double EPS = 1e-8;
const int MAXN = 5010;
int dx[] = {0, 1, 0, -1}, dy[] = {-1, 0, 1, 0};

struct sticks
{
    int l, w;
    bool flag;
    bool operator < (const sticks& b) const
    {
        if (l == b.l)
            return w < b.w;
        return l < b.l;
    }
};

sticks st[MAXN];

int main()
{
    int kase;
    cin >> kase;
    while (kase--)
    {
        int n;
        cin >> n;
        for (int i = 0; i < n; ++i)
        {
            scanf("%d%d", &st[i].l, &st[i].w);
            st[i].flag = false;
        }
        sort(st, st+n);
        int time = 0;
        for (int i = 0; i < n; ++i)
            if (!st[i].flag)
            {
                time++;
                st[i].flag = true;
                int w = st[i].w;
                for (int j = i+1; j < n; ++j)
                    if (!st[j].flag && st[j].w >= w)
                    {
                        st[j].flag = true;
                        w = st[j].w;
                    }
            }
        cout << time << endl;
    }
    return 0;
}