Introduction to Fuzzy sorting algorithm 7.6 C + + implementation code

The topic read a bit obscure, after understanding the thought is more easy to think. In fact, the order of the elements replaced by the closed interval, and the closed interval ' equal ' is that these intervals have a common area (equivalence class), so similar to exercise 7.2, return two subscript, they are equal elements, the front is the ' smaller ' interval, followed by the ' larger ' interval. So how to find the common sub-range, I here is the past to iterate over the array, if there is overlap, the main element is updated to overlapping areas, and then traverse, if the next interval and the main element overlap, then the interval and all the previous interval overlap.

In the second question, if n intervals have overlapping parts, that is, n elements are ' equal ', it is clear that only O (n) time is required, because the first division ends the sort.

#include <iostream>#include<algorithm>using namespaceStd;typedef pair<int,int>Interval;//determine if two intervals overlapBOOLIntersect (Interval i1, Interval i2) {returnI1.second >= I2.first&&i1.first <=I2.second;} Pair<int,int> interval_p (Interval *inv,intPintR) {    intPivot = p + rand ()% (R-p +1); Interval Temp=Inv[pivot];    Swap (Inv[pivot], inv[r]);  for(intK =0; K < R; ++k) {        if(Intersect (inv[k], temp)) {Temp.first=Max (Inv[k].first, Temp.first); Temp.second=min (Inv[k].second, temp.second); }    }    inti = P-1, j =p;  for(; J < R; + +)j) {        if(!Intersect (inv[j],temp))if(Inv[j].first <Temp.first) {                ++i;            Swap (Inv[j], inv[i]); }    }    intt = i+1;  for(j = i +1; J < R; ++j) {        if(Intersect (inv[j], temp)) {++i;        Swap (Inv[i], inv[j]); }} swap (Inv[i+1], inv[r]); return{T, i +1 };}//the input is an array subscriptvoidInterval_qs (Interval *inv,intPintR) {    if(P >=R)return; Pair<int,int> pa=interval_p (INV, P, R); Interval_qs (INV, p, Pa.first-1); Interval_qs (INV, Pa.second+1, R);}//Review the tail recursion, stack depth problemvoidTail_iqs (Interval *inv,intPintR) {     while(P <R) {Pair<int,int> PA =interval_p (INV, P, R); Tail_iqs (INV, p, Pa.first-1); P= Pa.second +1; }}intMain () {Interval inv[ A] = { {1,2}, {2,3}, {3,9}, {2,3}, {5,7}, {1,9}, {2,6}, {0,1}, {1,1}, {2,2}, {3,4}, {9,Ten} }; Tail_iqs (INV,0, One);  for(inti =0; I < A; ++i) cout<< Inv[i].first <<" "<< inv[i].second<<Endl;}

Introduction to Fuzzy sorting algorithm 7.6 C + + implementation code