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假设S[n]表示n条线段中最长重叠距离,最长重叠距离只与两条线段有关,那么考虑两种情况:
1. 最长重叠距离与第n条线段无关,则最长重叠距离存在于前n-1条线段中,即S[n]=S[n-1];
2. 最长重叠距离与第n条线段有关,则最长重叠距离为第n条线段与前面n-1条线段中的最大重叠距离者,S[n]=max(overlap(segment[n],segment[i])), i=1....n-1
因此得到状态转移方程:
S[n]=0; n<=1
S[2]=overlap(segment[0],segment[1])
S[n]=max(S[n-1],max(overlap(segment[n],segment[i])))
#include <iostream>
#include <algorithm>
using namespace std;
typedef struct{
int start;
int end;
}segment;
bool isShorter(const segment &s1,const segment &s2);
segment commonSegment(const segment &seg_a,const segment &seg_b);
int findLongestSegment(const vector<segment> &segments,int size,segment &maxSegment);
int main()
{
int n;
int start,end;
while(cin>>n){
vector<segment> segments(n);
for(int i=0;i<n;i++){
if(cin>>start && cin>>end){
segments[i].start=start;
segments[i].end=end;
}
}
// sort by segment.end
sort(segments.begin(),segments.end(),isShorter);
segment maxSeg;
int maxLen;
maxLen=findLongestSegment(segments,n,maxSeg);
cout<<"Longest Length:"<<maxLen<<endl;
cout<<"start:"<<maxSeg.start<<endl;
cout<<"end:"<<maxSeg.end<<endl;
}
return 0;
}
bool isShorter(const segment &s1,const segment &s2){
return s1.end<s2.end;
}
//assume seg_a.end<seg_b.end
segment commonSegment(const segment &seg_a,const segment &seg_b){
segment commonSeg;
if(seg_a.end<seg_b.start){
commonSeg.start=0;
commonSeg.end=0;
}
else{
commonSeg.start=seg_b.start;
commonSeg.end=seg_a.end;
}
return commonSeg;
}
int findLongestSegment(const vector<segment> &segments,int size,segment &maxSegment){
if(size<=0)
return 0;
/*
if(size==1){
maxSegment.start=segments[0].start;
maxSegment.end=segments[0].end;
return maxSegment.end-maxSegment.start;
}
*/
if(size==2){
maxSegment=commonSegment(segments[0],segments[1]);
return maxSegment.end-maxSegment.start;
}
// size>2
// dynamic programming
int maxLen,tmpLen;
segment tmpMaxSeg;
maxLen=findLongestSegment(segments,size-1,tmpMaxSeg);
maxSegment=tmpMaxSeg;
// maxSegment=(maxSegment,commonSeg)
segment commonSeg;
for(int i=0;i<size-1;i++){
commonSeg=commonSegment(segments[i],segments[size-1]);
tmpLen=commonSeg.end-commonSeg.start;
if(tmpLen>maxLen){
maxLen=tmpLen;
maxSegment=commonSeg;
}
}
return maxSegment.end-maxSegment.start;
}

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原文地址:http://www.cnblogs.com/AndyJee/p/4537376.html