http://acm.hdu.edu.cn/showproblem.php?pid=1051
Wooden Sticks
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9167 Accepted Submission(s): 3748
Problem 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
Source
Asia 2001, Taejon (South Korea)
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解析:
思路:
按照木棍从小到的顺序排好,其实就是求重量的最长上升子序列个数
78MS 288K 912 B
*/
#include<stdio.h> #include<string.h> #include<algorithm> #include <iostream> using namespace std; const int maxn=5010; int dp[maxn]; struct Node { int l; int w; }node[maxn]; bool cmp( Node a1, Node a2) { return a1.l>a2.l||(a1.l==a2.l&&a1.w>a2.w); } int main() {int T,i,n,ans,j; scanf("%d",&T); while(T--) { scanf("%d",&n); if(n==0) {printf("0\n"); continue; } for(i=0;i<n;i++) scanf("%d%d",&node[i].l,&node[i].w); sort(node,node+n,cmp); memset(dp,0,sizeof(dp)); dp[0]=1; ans=0; int m; for(i=1;i<n;i++) { m=dp[i]; for(j=0;j<i;j++) {if(node[i].w>node[j].w)//因为l已经排好序了,只需比较w即可 { if(dp[j]>m) m=dp[j]; } } dp[i]=m+1; if(ans<dp[i]) ans=dp[i]; } // for(i=0;i<n;i++) //printf("l=%d,w=%d\n",node[i].l,node[i].w); printf("%d\n",ans); } //system("pause"); return 0; }