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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。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <ctime> 6 using namespace std; 7 8 #define MAXN 100005 9 10 int a[MAXN], b[MAXN]; 11 int c[MAXN]; 12 int ans, n; 13 14 void merge(int *a, int l, int r) { 15 int mid = (l+r)>>1; 16 int i = l, j = mid+1; 17 int n = 0; 18 19 while (i<=mid && j<=r) { 20 if (a[i] <= a[j]) { 21 c[n++] = a[i++]; 22 } else { 23 c[n++] = a[j++]; 24 ans += (mid-i+1); 25 } 26 } 27 while (i <= mid) 28 c[n++] = a[i++]; 29 while (j <= r) 30 c[n++] = a[j++]; 31 for (i=0, j=l; i<n; ++i, ++j) 32 a[j] = c[i]; 33 } 34 35 void mergeSort(int *a, int l, int r) { 36 int mid = (l+r)>>1; 37 38 if (l >= r) 39 return ; 40 mergeSort(a, l, mid); 41 mergeSort(a, mid+1, r); 42 merge(a, l, r); 43 } 44 45 void bruteSolve(int *b, int l, int r) { 46 int i, j; 47 48 for (i=l; i<=r; ++i) { 49 for (j=l; j<=i; ++j) { 50 if (b[j] > b[i]) 51 ++ans; 52 } 53 } 54 } 55 56 void init() { 57 int i; 58 59 n = rand()%(MAXN-1)+1; 60 for (i=1; i<=n; ++i) { 61 a[i] = rand(); 62 b[i] = a[i]; 63 } 64 } 65 66 void solve() { 67 clock_t beg, end; 68 int tmp; 69 70 ans = 0; 71 beg = clock(); 72 mergeSort(a, 1, n); 73 end = clock(); 74 printf("nlgn: ans = %d\n", ans); 75 printf(" time = %.2lf\n", (double)(end-beg)/CLOCKS_PER_SEC); 76 77 tmp = ans; 78 ans = 0; 79 beg = clock(); 80 bruteSolve(b, 1, n); 81 end = clock(); 82 printf("n*n : ans = %d\n", ans); 83 printf(" time = %.2lf\n", (double)(end-beg)/CLOCKS_PER_SEC); 84 85 if (tmp != ans) 86 printf("**** wrong ****\n"); 87 printf("\n"); 88 } 89 90 int main() { 91 int t = 30; 92 93 while (t--) { 94 init(); 95 solve(); 96 } 97 98 return 0; 99 }
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原文地址:http://www.cnblogs.com/bombe1013/p/4231839.html