HDU 1051 Wooden Sticks(How to use 快排)
2014-12-07 19:52
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Wooden Sticks |
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) |
Total Submission(s): 2331 Accepted Submission(s): 908 |
[align=left]Problem Description[/align] 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). |
[align=left]Input[/align] 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. |
[align=left]Output[/align] The output should contain the minimum setup time in minutes, one per line. |
[align=left]Sample Input[/align]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 |
[align=left]Sample Output[/align]2 1 3 |
#include<stdio.h> #include<iostream> #include<string.h> #include<algorithm> //1.使用qsort()需 #include<algorithm> OR #include<stdlib.h> using namespace std; struct tim { int x,y; }a[11111]; int tab[50001]; int comp(const void *a,const void *b) //2.定义比较函数 { tim *x=(tim*)a,*y=(tim*)b; if(x->x!=y->x) return x->x-y->x; return x->y-y->y; } //再举个简单的栗子: /* int compare(const void *a,const void *b){ return *(int*)a - *(int*)b; //if return *(int*b) - *(int*a) ,then the array will be sortted in decreasing order (递减序列) } */ int main() { int i,j,m,n,k,t; cin>>k; while(k--) { cin>>n; for(i=1;i<=n;i++) cin>>a[i].x>>a[i].y; /*3. 实参1.传要排序的数组 +1表示从第一个开始排序 2.n 数组元素个数 3.一个元素的大小 4.比较函数 OK 快排的实现请参照 http://blog.csdn.net/u012400327/article/details/22619497 */ qsort(a+1, n, sizeof(tim), comp); memset(tab, 0, sizeof(tab)); m=j=0; while(j<n) { m++; i=1; while(tab[i]==1) i++; int temp1=i; tab[i]=1; j++; i++; // cout<<temp1<<' '<<i<<' '<<j<<endl; while(i<=n) if(tab[i]==0) { { if(a[i].x>=a[temp1].x&&a[i].y>=a[temp1].y) { temp1=i; tab[i]=1; j++; } } i++; } else i++; } cout<<m<<endl; } }
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