Wooden Sticks
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9125 Accepted Submission(s): 3729
Problem DescriptionThere 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).
InputThe 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.
OutputThe 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
SourceAsia 2001, Taejon (South Korea)
#include<stdio.h>#include<algorithm>using namespace std;struct node //儲存棍子屬性,長,重,是否訪問 { int l; int w; int v;}str[5002];bool cmp(node a,node b) //按長度、重量從小到大排序{ if(a.l==b.l) return a.w<b.w; return a.l<b.l;}int main(){ int t,n; int tim; int i,j; scanf("%d",&t); while(t--) { tim=0; scanf("%d",&n); for(i=0;i<n;i++) { scanf("%d%d",&str[i].l,&str[i].w); str[i].v=0; } sort(str,str+n,cmp); //排序 int l,w; for(i=0;i<n;i++) //選取第i根木棒(排序後)為加工時的第一根木棒 { if(str[i].v==0) { l=str[i].l; w=str[i].w; str[i].v=1; tim++; } for(j=i+1;j<n;j++) { if(str[j].v==0&&str[j].l>=l&&str[j].w>=w) { l=str[j].l; w=str[j].w; str[j].v=1; } } } printf("%d\n",tim); } return 0;}