【Algorithm】逆序数的分治求解

逆序数的分治求解，时间复杂度O(nlgn)。基本思想是在归并排序的基础上加逆序计数。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <ctime>
 using namespace std;

 #define MAXN 100005

 int a[MAXN], b[MAXN];
 int c[MAXN];
 int ans, n;

 void merge(int *a, int l, int r) {
     int mid = (l+r)>>;
     int i = l, j = mid+;
     int n = ;

     while (i<=mid && j<=r) {
         if (a[i] <= a[j]) {
             c[n++] = a[i++];
         } else {
             c[n++] = a[j++];
             ans += (mid-i+);
         }
     }
     while (i <= mid)
         c[n++] = a[i++];
     while (j <= r)
         c[n++] = a[j++];
     for (i=, j=l; i<n; ++i, ++j)
         a[j] = c[i];
 }

 void mergeSort(int *a, int l, int r) {
     int mid = (l+r)>>;

     if (l >= r)
         return ;
     mergeSort(a, l, mid);
     mergeSort(a, mid+, r);
     merge(a, l, r);
 }

 void bruteSolve(int *b, int l, int r) {
     int i, j;

     for (i=l; i<=r; ++i) {
         for (j=l; j<=i; ++j) {
             if (b[j] > b[i])
                 ++ans;
         }
     }
 }

 void init() {
     int i;

     n = rand()%(MAXN-)+;
     for (i=; i<=n; ++i) {
         a[i] = rand();
         b[i] = a[i];
     }
 }

 void solve() {
     clock_t beg, end;
     int tmp;

     ans = ;
     beg = clock();
     mergeSort(a, , n);
     end = clock();
     printf("nlgn: ans  = %d\n", ans);
     printf("      time = %.2lf\n", (double)(end-beg)/CLOCKS_PER_SEC);

     tmp = ans;
     ans = ;
     beg = clock();
     bruteSolve(b, , n);
     end = clock();
     printf("n*n : ans  = %d\n", ans);
     printf("      time = %.2lf\n", (double)(end-beg)/CLOCKS_PER_SEC);

     if (tmp != ans)
         printf("**** wrong ****\n");
     printf("\n");
 }

 int main() {
     int t = ;

     while (t--) {
         init();
         solve();
     }

     return ;
 }