1065 Wooden Sticks
題目描述:連結
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
題目理解:
先輸入T,代表輸入案列。再輸入n,代表接下來會有n對資料。每一對代表木棒的起點與終點。
有一個機器處理這些木棒,機器開啟的時候需要耗費1分鐘,如果第i+1個木棒的重量和長度都大於等於第i個處理的木棒,則不會耗費時間,否則需要消耗1分鐘。
問加工這些木棒所需要的最小消耗時間。
思路:
先對木棒進行排序,起點遞增,若起點相同則終點遞增。
程式碼:
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
struct node
{
int l,w;
}s[5010];
bool flag[5010];
bool cmp(node a,node b) //將木棒排序 起點遞增 若起點相同則終點遞增
{
if(a.l!=b.l)
return a.l<b.l;
return a.w<b.w;
}
int main()
{
int T, n;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%d%d",&s[i].l,&s[i].w);
sort(s,s+n,cmp);
int sum=0;
int maxn;
memset(flag,0,sizeof(flag));
for(int i=0;i<n;i++)//對第i個木棒求遞增子序列
{
maxn=s[i].w;//記錄新序列的開始
if(!flag[i])//木棒未曾用過
{
for(int j=i+1;j<n;j++)
if(maxn<=s[j].w&&!flag[j])//又有了更大的值&&未曾用過的
{
flag[j]=1;//標記用過
maxn=s[j].w;//重新整理最大值
}
sum++;//時間+1
}
}
cout<<sum<<endl;
}
return 0;
}