POJ 1065 Wooden Sticks

Wooden Sticks

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 16372 Accepted: 6794

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

 

题意：

将不同长度和重量的木棍尽可能按两个属性的不下降序列排列，求其排列中的倒序数

 

代码：

#include<cstdio>#include<cstring>#include<cstdlib>#define MAX 5002typedef struct Stick{int l,w;};Stick st[MAX];int t,n;int cmp(const void *a,const void *b){Stick *c=(Stick *)a,*d=(Stick *)b;if(c->l!=d->l) return c->l-d->l;else return c->w-d->w;}int main(){int i,j,min;while(scanf("%d",&t)!=EOF){while(t--){scanf("%d",&n);for(i=0;i<n;i++)scanf("%d%d",&st[i].l,&st[i].w);qsort(st,n,sizeof(struct Stick),cmp);min=0;for(i=0;i<n;i++){int tmp=st[i].w;if(st[i].w){min++;for(j=i+1;j<n;j++){if(st[j].w>=tmp){tmp=st[j].w;st[j].w=0;}}}}printf("%d\n",min);}}return 0;}

思路：

求最少时间，本质为将一个属性按非降序排列后，看另一个属性排列尽可能少的倒序数。