(演算法)最長重疊線段或區間

標籤：

題目：X軸上有N條線段，每條線段包括1個起點和終點。線段的重疊是這樣來算的，[10 20]和[12 25]的重疊部分為[12 20]。給出N條線段的起點和終點，從中選出2條線段，這兩條線段的重疊部分是最長的。輸出這個最長的距離。如果沒有重疊，輸出0。 思路：1、暴力計算依次計算兩兩線段之間的重疊長度，但複雜度太高2、動態規劃

假設S[n]表示n條線段中最長重疊距離，最長重疊距離只與兩條線段有關，那麼考慮兩種情況：

1. 最長重疊距離與第n條線段無關，則最長重疊距離存在於前n-1條線段中，即S[n]=S[n-1]；

2. 最長重疊距離與第n條線段有關，則最長重疊距離為第n條線段與前面n-1條線段中的最大重疊距離者，S[n]=max(overlap(segment[n],segment[i])), i=1....n-1

因此得到狀態轉移方程：

S[n]=0; n<=1

S[2]=overlap(segment[0],segment[1])

S[n]=max(S[n-1],max(overlap(segment[n],segment[i])))

代碼：

#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.endsegment 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;}

運行結果： 

(演算法)最長重疊線段或區間