zoj1025贪心算法

Wooden Sticks Time Limit: 1 Second      Memory Limit: 32768 KB

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

Output for the Sample Input

2
1
3 

代码：

#include <iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>

using namespace std;
const int maxN = 5001;
struct stick{
    int l;//木棒的长度
    int w;//木棒的重量
};

int cmp(const void *a,const void *b)
{
    stick *sa = (stick *)a;
    stick *sb = (stick *)b;
    if(sa->l==sb->l)//长度相等时，按重量排序
        return sb->w - sa->w;
    return sb->l-sa->l;//优先按长度排序
}
//计算重量w的最长单调递增子序列的个数
int LIS(int n,stick a[])
{
    int b[maxN];//存放子结构的最长递增子序列的长度
    memset(b,0,sizeof(int));
    int i,j,k;
    b[0]=1;
    for(i=1;i<n;i++)
    {
        //计算到第i个元素的最长递增子序列的长度
        k=0;
        for(j=0;j<i;j++)
            if(a[j].w < a[i].w&&k<b[j])
            k=b[j];
            b[i]=k+1;
    }
    //求数组中b[]的最大值
    int max=0;
    for(i=0;i<n;i++)
        if(b[i]>max)
            max = b[i];
        return max;
}
int main()
{
    stick data[maxN];
    int i,k;
    int test;
    scanf("%d",&test);
    for(k=0;k<test;++k)
    {
        int n;
        scanf("%d",&n);
        for(i=0;i<n;++i)
        {
            scanf("%d%d",&data[i].l,&data[i].w);
        }
        qsort(data,n,sizeof(stick),cmp);
        printf("%d\n",LIS(n,data));

    }
    return 0;
}