poj_1065_Wooden Sticks

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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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 

 

按照棍子长度从小到大排序，之后从前向后遍历，找到每一个都比前一个l，w都大的位置，把它们相应的位置信息都置为1。然后再从前到后扫描，看没有置为1的w加和。

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


using namespace std;


struct node
{
    int l,w;
};
int vis[5002]={0};
node a[5002];
bool cmp(node a,node b)
{
    return a.l<b.l||(a.l==b.l&&a.w<b.w);
}
void solve(int n)
{
    int cnt=0;
    for(int i=0;i<n;i++)
        {
            if(!vis[i])
            {
                vis[i] = 1;
                cnt++;
                int w = a[i].w;
                for(int j = i+1; j < n; j++)
                {
                    if(!vis[j] && a[j].w >= w)
                    {
                        vis[j] = 1;
                        w = a[j].w;
                    }
                }
            }
        }
        printf("%d\n",cnt);
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        memset(vis,0,sizeof(vis));
        for(int i=0;i<n;i++) scanf("%d%d",&a[i].l,&a[i].w);
        sort(a,a+n,cmp);
        solve(n);
    }
    return 0;
}