Wooden Sticks（hdu1501）（sort，dp）

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 9541    Accepted Submission(s): 3917

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

 

好吧，这是一道简单题，，，仔细点就好；

思路是；把木棍按照长度优先从小到大排好，当长度一定是按宽度从小到大排好，然后比较宽度就ok了

，具体见代码，，，，

 

简单易懂型：

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
struct node
{
    int l;
    int w;
    int k;
}a[5005];

int main()
{
    int T,n,j,i,temp,sum,p;
    scanf("%d",&T);
    while(T--)
    {
        sum=0;
        scanf("%d",&n);
        for(i=1;i<=n;i++)
        {
            scanf("%d%d",&a[i].l,&a[i].w);
            a[i].k=1;//标记
        }
        for(i=1;i<=n;i++)//选择                     why!!!
        {
            p=i;
            for(j=i+1;j<=n;j++)
                if(a[p].l>a[j].l||a[p].l==a[j].l&&a[p].w>a[j].w) //长度优先排，然后是宽度，都排好，从小到大
                    p=j;
            if(p!=i)
            {
                int t;
                t=a[i].l;
                a[i].l=a[p].l;
                a[p].l=t;
                t=a[i].w;
                a[i].w=a[p].w;
                a[p].w=t;
            }
        }
//        for(i=1;i<=n;i++)
//            printf("%d %d %d \n",a[i].l,a[i].w,a[i].k);
        for(i=1;i<=n;i++)
        {
            if(a[i].k==1)
            {
                temp=a[i].w;
                for(j=i+1;j<=n;j++)
                {
                    if(a[j].w>=temp&&a[j].k==1)
                    {
                        a[j].k=0;
                        temp=a[j].w;
                        sum++;
                    }

                }
            }
        }
        printf("%d\n",n-sum);
    }
    return 0;
}

/*
4
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
7
1 12 1 5 2 6 2 8 2 4 4 3 3 7

*/

然后是更加简洁的排序sort，快排。。。

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
struct node
{
    int l;
    int w;
    int k;
}a[5005];

int cmp(node a,node b)
{
    if(a.w==b.w)
        return a.l<b.l;
    return a.w<b.w;
}


int main()
{
    int T,n,j,i;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            scanf("%d%d",&a[i].l,&a[i].w);
            a[i].k=1;//标记
        }
        sort(a,a+n,cmp);
        for(i=0;i<n;i++)
            printf("%d %d %d \n",a[i].l,a[i].w,a[i].k);

        int num=n;
        int count=0;
        while(num)
        {
            int tl=0;int tw=0;
            for(i=0;i<n;i++)
            {
                if(a[i].k==1&&a[i].w>=tw&&a[i].l>=tl)
                {
                    tl=a[i].l; tw=a[i].w; a[i].k=0;num--;
                }
            }
            count++;
        }
        printf("%d\n",count);
    }
    return 0;
}

稍作修改：

 

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
struct node
{
    int l;
    int w;
    int k;
}a[5005];

int cmp(node a,node b)//从小到大排，，，对于结构体
{
    if(a.l==b.l)
        return a.w<b.w;
    return a.l<b.l;
}
int main()
{
    int T,n,j,i,temp,sum;
    scanf("%d",&T);
    while(T--)
    {
        sum=0;
        scanf("%d",&n);
        for(i=1;i<=n;i++)
        {
            scanf("%d%d",&a[i].l,&a[i].w);
            a[i].k=1;//标记
        }
        sort(a+1,a+n+1,cmp);
        for(i=1;i<=n;i++)
            printf("%d %d %d \n",a[i].l,a[i].w,a[i].k);
        for(i=1;i<=n;i++)
        {
            if(a[i].k==1)
            {
                temp=a[i].w;
                for(j=i+1; j<=n; j++)
                {
                    if(a[j].w>=temp&&a[j].k==1)
                    {
                        a[j].k=0;
                        temp=a[j].w;
                        sum++;
                    }

                }
            }
        }
        printf("%d\n",n-sum);
    }
    return 0;
}

是的，冒泡嘛，也是可以的，，，

 

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
struct node
{
    int l;
    int w;
    int k;
}a[5005];

int cmp(node a,node b)
{
    if(a.l==b.l)
        return a.w<b.w;
    return a.l<b.l;
}
int main()
{
    int T,n,j,i,temp,sum;
    scanf("%d",&T);
    while(T--)
    {
        sum=0;
        scanf("%d",&n);
        for(i=1;i<=n;i++)
        {
            scanf("%d%d",&a[i].l,&a[i].w);
            a[i].k=1;//标记
        }
        for(i=1;i<=n;i++)//冒泡
        {
            for(j=0;j<n-i;j++)//j从1开始就错了。。。
                if(a[j].l>a[j+1].l)
                    {
                        int vv;
                        vv=a[j].w; a[j].w=a[j+1].w; a[j+1].w=vv;
                        vv=a[j].l; a[j].l=a[j+1].l; a[j+1].l=vv;
                    
                    }
        }

        for(i=1;i<=n;i++)
            printf("%d %d %d \n",a[i].l,a[i].w,a[i].k);
        for(i=1;i<=n;i++)
        {
            if(a[i].k==1)
            {
                temp=a[i].w;
                for(j=i+1; j<=n; j++)
                {
                    if(a[j].w>=temp&&a[j].k==1)
                    {
                        a[j].k=0;
                        temp=a[j].w;
                        sum++;
                    }

                }
            }
        }
        printf("%d\n",n-sum);
    }
    return 0;
}

 

推荐第二种，，，，本题还可以用dp做，具体代码，有待有序更新。。。o(∩_∩)o 哈哈