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http://poj.org/problem?id=2299
In this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. For the input sequence
Ultra-QuickSort produces the output
Your task is to determine how many swap operations Ultra-QuickSort needs to perform in order to sort a given input sequence.
The input contains several test cases. Every test case begins with a line that contains a single integer n < 500,000 -- the length of the input sequence. Each of the the following n lines contains a single integer 0 ≤ a[i] ≤ 999,999,999, the i-th input sequence element. Input is terminated by a sequence of length n = 0. This sequence must not be processed.
For every input sequence, your program prints a single line containing an integer number op, the minimum number of swap operations necessary to sort the given input sequence.
5
9
1
0
5
4
3
1
2
3
0
6
0
1 #include<algorithm> 2 #include<iostream> 3 #include<cstdlib> 4 #include<cstring> 5 #include<cstdio> 6 #include<vector> 7 #include<map> 8 #include<set> 9 using std::cin; 10 using std::cout; 11 using std::endl; 12 using std::find; 13 using std::sort; 14 using std::set; 15 using std::map; 16 using std::pair; 17 using std::vector; 18 using std::multiset; 19 using std::multimap; 20 #define all(c) (c).begin(), (c).end() 21 #define iter(c) decltype((c).begin()) 22 #define cls(arr,val) memset(arr,val,sizeof(arr)) 23 #define cpresent(c, e) (find(all(c), (e)) != (c).end()) 24 #define rep(i, n) for (int i = 0; i < (int)(n); i++) 25 #define tr(c, i) for (iter(c) i = (c).begin(); i != (c).end(); ++i) 26 #define pb(e) push_back(e) 27 #define mp(a, b) make_pair(a, b) 28 #define lc (root<<1) 29 #define rc (root<<1|1) 30 #define mid ((l+r)>>1) 31 const int Max_N = 500100; 32 typedef unsigned long long ull; 33 struct Node { int val; }; 34 struct P { 35 int v, id; 36 friend bool operator<(const P &a, const P &b) { 37 return a.v == b.v ? a.id < b.id : a.v < b.v; 38 } 39 }rec[Max_N]; 40 struct SegTree { 41 Node seg[Max_N << 2]; 42 inline void init() { 43 cls(seg, 0); 44 } 45 inline void insert(int root, int l, int r, int p) { 46 if (p > r || p < l) return; 47 if (p <= l && p >= r) { seg[root].val++; return; } 48 insert(lc, l, mid, p); 49 insert(rc, mid + 1, r, p); 50 seg[root].val = seg[lc].val + seg[rc].val; 51 } 52 inline ull query(int root, int l, int r, int x, int y) { 53 if (x > r || y < l) return 0; 54 if (x <= l && y >= r) return seg[root].val; 55 ull ret = 0; 56 ret += query(lc, l, mid, x, y); 57 ret += query(rc, mid + 1, r, x, y); 58 return ret; 59 } 60 }seg; 61 int main() { 62 #ifdef LOCAL 63 freopen("in.txt", "r", stdin); 64 freopen("out.txt", "w+", stdout); 65 #endif 66 int n; 67 while (~scanf("%d", &n) && n) { 68 seg.init(); 69 ull res = 0; 70 for (int i = 1; i <= n; i++) scanf("%d", &rec[i].v), rec[i].id = i; 71 sort(rec + 1, rec + n + 1); 72 for (int i = 1; i <= n; i++) { 73 res += seg.query(1, 1, n, rec[i].id, n); 74 seg.insert(1, 1, n, rec[i].id); 75 } 76 printf("%lld\n", res); 77 } 78 return 0; 79 }
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原文地址：http://www.cnblogs.com/GadyPu/p/4572450.html
