POJ1065 Wooden Sticks

Wooden Sticks

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 15343 Accepted: 6328

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

213

Source

Taejon 2001

这是一个贪心问题，好吧，我承认在理解题意上花了好多时间啊，wa了好多次，就是确定最少多少组非降序数组

#include<iostream>#include<algorithm>#include<cstring>#include<cstdio>using namespace std;struct node{    int x,y,z;//z记录是否已经处理}p[5001];bool pai(node a,node b)//排序{    if(a.x==b.x)return a.y<b.y;    else return a.x<b.x;}int main(){    int t,n;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(int i=0;i<n;i++)        {            scanf("%d%d",&p[i].x,&p[i].y);            p[i].z=0;        }        sort(p,p+n,pai);        int sum=0;        for(int i=1;i<n;i++)        {            if(p[i].z==0)            {                sum++;                int temp=p[i].y;//排除处理了i后不用处理的棒子                for(int j=i+1;j<n;j++)                {                    if(p[j].y>=temp&&p[j].z==0)                    {                        p[j].z=1;                        temp=p[j].y;                    }                }            }        }        cout<<sum<<endl;    }    return 0;}