[hdu4911]逆序对相关
思路：由于只能交换相邻的数，所以每次最多减小1个逆序对（且如果存在逆序对那么肯定可以减小1个）！于是乎。。就是统计逆序对的裸题了。树状数组或归并都行。
1 #pragma comment(linker, "/STACK:10240000,10240000") 2 3 #include <iostream> 4 #include <cstdio> 5 #include <algorithm> 6 #include <cstdlib> 7 #include <cstring> 8 #include <map> 9 #include <queue> 10 #include <deque> 11 #include <cmath> 12 #include <vector> 13 #include <ctime> 14 #include <cctype> 15 #include <set> 16 17 using namespace std; 18 19 #define mem0(a) memset(a, 0, sizeof(a)) 20 #define lson l, m, rt << 1 21 #define rson m + 1, r, rt << 1 | 1 22 #define define_m int m = (l + r) >> 1 23 #define Rep(a, b) for(int a = 0; a < b; a++) 24 #define lowbit(x) ((x) & (-(x))) 25 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {} 26 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {} 27 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {} 28 29 typedef double db; 30 typedef long long LL; 31 typedef pair<int, int> pii; 32 typedef multiset<int> msi; 33 typedef multiset<int>::iterator msii; 34 typedef set<int> si; 35 typedef set<int>::iterator sii; 36 typedef vector<int> vi; 37 38 const int dx[8] = {1, 0, -1, 0, 1, 1, -1, -1}; 39 const int dy[8] = {0, -1, 0, 1, -1, 1, 1, -1}; 40 const int maxn = 1e5 + 7; 41 const int maxm = 1e5 + 7; 42 const int maxv = 1e7 + 7; 43 const int MD = 1e9 +7; 44 const int INF = 1e9 + 7; 45 const double PI = acos(-1.0); 46 const double eps = 1e-10; 47 48 int tmp[maxn], a[maxn]; 49 50 LL merge(int l, int m, int r) { 51 int p = m + 1, t = l; 52 LL res = 0; 53 for (int i = l; i <= m; i++) { 54 while (p <= r && a[i] > a[p]) { 55 tmp[t++] = a[p++]; 56 } 57 res += p - m - 1; 58 tmp[t++] = a[i]; 59 } 60 for (int i = l; i < p; i++) a[i] = tmp[i]; 61 return res; 62 } 63 64 LL merge_sort(int l, int r) { 65 if (l == r) return 0; 66 define_m; 67 return merge_sort(l, m) + merge_sort(m + 1, r) + merge(l, m, r); 68 } 69 70 int main() { 71 //freopen("in.txt", "r", stdin); 72 int n; 73 LL k; 74 while (cin >> n >> k) { 75 for (int i = 0; i < n; i++) { 76 scanf("%d", a + i); 77 } 78 LL ans = merge_sort(0, n - 1) - k; 79 ans = max(ans, 0LL); 80 cout << ans << endl; 81 } 82 return 0; 83 }
Hint:做这题的时候发现一个不容易发现的坑（与这题本身无关），多个函数相加并不一定按从左至右的顺序执行（可能是为了某种意义上地优化代码）。此代码用c++交好像w会a掉，不信可以一试~(c++貌似优化程度比较高？)
