ZOJ Problem Set - 1076 Gene Assembly

ZOJ Problem Set - 1076

Gene Assembly

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Time Limit: 2 Seconds                                    Memory Limit: 65536 KB                            

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Statement of the Problem

With the large amount of genomic DNA sequence data being made available, it   is becoming more important to find genes (parts of the genomic DNA which are   responsible for the synthesis of proteins) in these sequences. It is known that   for eukaryotes (in contrast to prokaryotes) the process is more complicated,   because of the presence of junk DNA that interrupts the coding region of genes   in the genomic sequence. That is, a gene is composed by several pieces (called   exons) of coding regions. It is known that the order of the exons is maintained   in the protein synthesis process, but the number of exons and their lengths   can be arbitrary.

Most gene finding algorithms have two steps: in the first they search for possible   exons; in the second they try to assemble a largest possible gene, by finding   a chain with the largest possible number of exons. This chain must obey the   order in which the exons appear in the genomic sequence. We say that exon i   appears before exon j if the end of i precedes the beginning of j.

The objective of this problem is, given a set of possible exons, to find the   chain with the largest possible number of exons that cound be assembled to generate   a gene.

Input Format

Several input instances are given. Each instance begins with the number 0 <   n < 1000 of possible exons in the sequence. Then, each of the next n lines   contains a pair of integer numbers that represent the position in which the   exon starts and ends in the genomic sequence. You can suppose that the genomic   sequence has at most 50000 basis. The input ends with a line with a single 0.

Output Format

For each input instance your program should print in one line the chain with   the largest possible number of exons, by enumerating the exons in the chain.   If there is more than one chain with the same number of exons, your program   can print anyone of them.

Sample Input

6
  340 500
  220 470
  100 300
  880 943
  525 556
  612 776
  3
  705 773
  124 337
  453 665
  0

Sample Output

3 1 5 6 4
  2 3 1

其实用贪心也可以做不过想练练动规就动规好了

AC代码：

#include<iostream>#include<algorithm>#include<string.h>using namespace std;struct Gene{int start,end,id;};bool cmp(Gene a,Gene b){if(a.start!=b.start){return a.start<b.start;}else return a.end<b.end;}void print(int *path,Gene*g,int start,int lay){if(start==-1)return;print(path,g,path[start],lay+1);cout<<g[start].id;if(lay!=0)cout<<' ';else cout<<endl;} int main(){int n;while(cin>>n&&n!=0){Gene g[1001];int path[1001];int count[1001];memset(path,-1,sizeof(path));memset(count,0,sizeof(count));for(int i=0;i<n;i++){cin>>g[i].start>>g[i].end;g[i].id=i+1; }sort(g,g+n,cmp);count[0]=1;for(int i=1;i<n;i++){for(int j=0;j<i;j++){if(g[j].end<g[i].start){if(count[i]<count[j]){count[i]=count[j];path[i]=j;}}}count[i]++; }int max=0;int flag;for(int i=0;i<n;i++)    if(max<count[i])       {       max=count[i];       flag=i;       }   //cout<<max<<endl;   print(path,g,flag,0);}}