http://acm.hdu.edu.cn/showproblem.php?pid=5792
World is Exploding
Problem Description
Given a sequence A with length n,count how many quadruple (a,b,c,d) satisfies: a≠b≠c≠d,1≤a<b≤n,1≤c<d≤n,Aa<Ab,Ac>Ad.
Input
The input consists of multiple test cases.
Each test case begin with an integer n in a single line.
The next line contains n integers A1,A2⋯An.
1≤n≤50000
0≤Ai≤1e9
Each test case begin with an integer n in a single line.
The next line contains n integers A1,A2⋯An.
1≤n≤50000
0≤Ai≤1e9
Output
For each test case,output a line contains an integer.
Sample Input
4
2 4 1 3
4
1 2 3 4
Sample Output
1
0
Insipired by http://blog.csdn.net/libin66/article/details/52098019
题意：比较简单直接看。
思路：分别求出在 i 的时候前面有多少比它小的（即像Aa，Ab这样升序的有多少对），用up记录，后面有多少比它小的（像Ac，Ad这样降序的有多少对），用down记录，然后ans = up * down，因为有很多重复。用树状数组维护四个数组，ls —— 在[1,i-1]有多少比a[i]小的，rs —— 在[i+1,n]有多少比a[i]小的，lb —— 在[1,i-1]有多少比a[i]大的，rb —— 在[i+1,n]有多少比a[i]大的。由容斥原理要减去 ( ls * lb + rs * rb + ls * rs + lb * rb )。
1 #include <cstdio> 2 #include <algorithm> 3 #include <cstring> 4 using namespace std; 5 #define N 50005 6 #define MOD 1000000007 7 typedef long long LL; 8 LL ls[N], lb[N], rs[N], rb[N], bit[N]; 9 int a[N], num[N], n, m; 10 11 /* 12 ls和lb是[1,i-1]中比a[i]小的数的数量，比a[i]大的数的数量 13 rs和rb是[i+1, n]中比a[i]小的数的数量，比a[i]大的数的数量 14 这些都可以通过树状数组实现 15 由于答案会多算进去a=c || a=d || b=c || b=d的情况，那么枚举这四种情况减去就可以了(a=c那么必定b!=d，同理其他 16 a=c：ans-=rs[i]*rb[i] 17 a=d：ans-=lb[i]*rb[i] 18 b=c：ans-=ls[i]*rs[i] 19 b=d：ans-=lb[i]*ls[i] 20 */ 21 22 void init() 23 { 24 memset(ls, 0, sizeof(ls)); 25 memset(lb, 0, sizeof(lb)); 26 memset(rs, 0, sizeof(rs)); 27 memset(rb, 0, sizeof(rb)); 28 memset(sum, 0, sizeof(sum)); 29 } 30 31 int lowbit(int x) 32 { 33 return x & (-x); 34 } 35 36 void update(int pos) 37 { 38 while(pos <= m) { 39 bit[pos]++; 40 pos += lowbit(pos); 41 } 42 } 43 44 LL query(int pos) 45 { 46 LL ans = 0; 47 while(pos > 0) { 48 ans += bit[pos]; 49 pos -= lowbit(pos); 50 } 51 return ans; 52 } 53 54 int main() 55 { 56 while(~scanf("%d", &n)) { 57 init(); 58 for(int i = 1; i <= n; i++) { 59 scanf("%d", &a[i]); 60 num[i] = a[i]; 61 } 62 sort(a+1, a+1+n); 63 m = unique(a+1, a+1+n) - (a+1); 64 for(int i = 1; i <= n; i++) 65 num[i] = lower_bound(a+1, a+1+m, num[i]) - a; 66 67 LL up = 0, down = 0; 68 memset(bit, 0, sizeof(bit)); 69 for(int i = 1; i <= n; i++) { 70 ls[i] = query(num[i] - 1); 71 lb[i] = query(m) - query(num[i]); 72 up += ls[i]; 73 update(num[i]); 74 } 75 76 memset(bit, 0, sizeof(bit)); 77 for(int i = n; i >= 1; i--) { 78 rs[i] = query(num[i] - 1); 79 rb[i] = query(m) - query(num[i]); 80 down += rs[i]; 81 update(num[i]); 82 } 83 84 LL ans = 0; 85 86 for(int i = 1; i <= n; i++) { 87 ans -= ls[i] * lb[i]; 88 ans -= rs[i] * ls[i]; 89 ans -= rb[i] * lb[i]; 90 ans -= rb[i] * rs[i]; 91 } 92 93 printf("%I64d ", up * down + ans); 94 } 95 return 0; 96 }