ZOJ 1076 Gene Assembly

Gene Assembly
Time Limit: 2 Seconds      Memory Limit: 65536 KB
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

把每对数看作线段坐标，那么题意就是要找不重合的最大线段个数。刚开始用开始的坐标排序，绕了弯路。用结束的坐标排序就简单多了。不过代码还是很丑，应该用struct，不过懒得改了。

#include <iostream>
#define MAX 1001

using namespace std;

bool is_bigger(int* inst1, int* inst2){
if(inst1[1] != inst2[1])
return (inst1[1] > inst2[1]);
else
return (inst1[0] > inst2[0]);
}

int main(){
int n_inst, temp, inst[MAX][3], sorted_inst[MAX][3], mark[MAX], end; //inst: (start pos, end pos, index)
cin>>n_inst;
while(n_inst > 0){
for(int i = 0; i < n_inst; i++){
cin>>inst[i][0]>>inst[i][1];
inst[i][2] = i + 1;
}
for(int i = 0; i < n_inst; i++){ // 排序
int count = 0;
for(int j = n_inst - 1; j >= 0; j--){
if(is_bigger(inst[i], inst[j]))
count++;
}
mark[i] = count;
}
for(int i = 0; i < n_inst; i++){
sorted_inst[mark[i]][0] = inst[i][0];
sorted_inst[mark[i]][1] = inst[i][1];
sorted_inst[mark[i]][2] = inst[i][2];
}
// cout<<endl;
// for(int i = 0; i < n_inst; i++)
// cout<<sorted_inst[i][2]<<' '<<sorted_inst[i][0]<<' '<<sorted_inst[i][1]<<endl;
cout<<sorted_inst[0][2];
end = sorted_inst[0][1];
for(int i = 1; i < n_inst; i++){
if(sorted_inst[i][0] > end){
cout<<' '<<sorted_inst[i][2];
end = sorted_inst[i][1];
}
}
cout<<endl;
cin>>n_inst;
}

return 0;
}