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Wooden Sticks

发布时间:2016-12-5 8:31:18 编辑:www.fx114.net 分享查询网我要评论
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E - Wooden Sticks Time Limit:1000MS    Memory Limit:32768KB    64bit IO Format:%I64d & %I64u SubmitStatus 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 2 1 3 #include<stdio.h> #include<string.h> #include<iostream> #include<algorithm> using namespace std; int main() { int n,m,l[5005],w[5005], t[5005],u[5005],d; int e[5005],r=0; scanf("%d",&m); while(m--) { memset(t,0,sizeof(t)); memset(u,0,sizeof(u)); memset(l,0,sizeof(l)); memset(w,0,sizeof(w)); scanf("%d",&n); for(int b=0;b<n;b++) scanf("%d%d",&l[b],&w[b]);//cin>>l[b]>>w[b]; for(int j=0;j<n-1;j++) for(int k=j+1;k<n;k++) { if(l[j]>l[k]) { d=l[j];l[j]=l[k];l[k]=d; d=w[j];w[j]=w[k];w[k]=d; } else if(l[j]==l[k]) { if(w[j]>w[k]) { d=w[j];w[j]=w[k];w[k]=d; } } } int count,x=0 ; for(int i=0;i<n;i++) { for(int j=0;j<=x;j++) { count =0; if(t[j]<=l[i]&&u[j]<=w[i]) {t[j]=l[i];u[j]=w[i];count=1;break;} } if(count ==0) { x++;t[x]=l[i];u[x]=w[i]; } } if(n==0) x=-1; e[r]=x+1; r++; } for(int i=0;i<r;i++) printf("%d\n",e[i]); return 0; }

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