BZOJ3289 Mato的文件管理

好烦的题，做了我2h，看来蒟蒻就是弱啊。。。

感觉就是莫队算法来着，然后就走上了乱搞的不归路、、、

莫队算法详情请上百度搜索，谢谢!>.<

这道题要求逆序对，所以可以用树状数组动态维护。每次转移复杂度就是树状数组点修改的复杂度，但是转移的时候好烦。。。（一定是我太弱了）

再加上分块，总复杂度为O(n * sqrt(n) * log(n))，刚刚好卡过去。。。

  1 /**************************************************************
  2     Problem: 3289
  3     User: rausen
  4     Language: C++
  5     Result: Accepted
  6     Time:8772 ms
  7     Memory:10200 kb
  8 ****************************************************************/
  9
 10 #include <cstdio>
 11 #include <cmath>
 12 #include <algorithm>
 13
 14 using namespace std;
 15 typedef long long LL;
 16
 17 struct data{
 18     int l, r, wl, w;
 19     LL ans;
 20 } a[150000];
 21 inline bool cmp(data a, data b){
 22     return a.wl == b.wl ? a.r < b.r : a.l < b.l;
 23 }
 24
 25 inline bool cmp_id(data a, data b){
 26     return a.w < b.w;
 27 }
 28
 29 LL x[150000], y[150000];
 30 LL BIT[150000], rank[150000];
 31 int TOT, T, tot, w[150000], block[150000];
 32 int n, Q, l, r, num;
 33 LL sum;
 34
 35 void build(){
 36     T = (int) sqrt(n), tot = 0;
 37     int x = 0;
 38     while (x < n)
 39         x += T, w[++tot] = x;
 40     w[tot] = min(w[tot], n);
 41     int y = 1;
 42     for (int i = 1; i <= n; ++i)
 43         if (i <= w[y]) block[i] = y;
 44         else block[i] = ++y;
 45 }
 46
 47 inline int lowbit(int x){
 48     return x & (-x);
 49 }
 50
 51 LL query(int x){
 52     if (x < 0) return 0;
 53     LL res = 0;
 54     while (x)
 55         res += BIT[x], x -= lowbit(x);
 56     return res;
 57 }
 58
 59 void change(int x, int del){
 60     while (x <= n)
 61         BIT[x] += del, x += lowbit(x);
 62 }
 63
 64 void update_right(int p, int del){
 65     sum += (LL) del * (num - query(x[p])), num += del;
 66     change(x[p], del);
 67 }
 68
 69 void update_left(int p, int del){
 70     change(x[p], del);
 71     sum += (LL) del * query(x[p] - 1), num += del;
 72 }
 73
 74 int find(int x){
 75     int l = 1, r = TOT, mid;
 76     for (; l != r; ){
 77         mid = (l + r) >> 1;
 78         if (rank[mid] >= x) r = mid;
 79         else l = mid + 1;
 80     }
 81     return l;
 82 }
 83
 84 int main(){
 85     scanf("%d", &n);
 86     for (int i = 1; i <= n; ++i){
 87         scanf("%lld", x + i);
 88         y[i] = x[i];
 89     }
 90     build();
 91     sort(y + 1, y + n + 1);
 92     rank[++TOT] = y[1];
 93     for (int i = 2; i <= n; ++i)
 94         if (y[i] != y[i - 1])
 95             rank[++TOT] = y[i];
 96     for (int i = 1; i <= n; ++i)
 97         x[i] = find(x[i]);
 98
 99     scanf("%d", &Q);
100     for (int i = 1; i <= Q; ++i){
101         scanf("%d%d", &a[i].l, &a[i].r);
102         a[i].wl = block[a[i].l], a[i].w = i;
103     }
104     sort(a + 1, a + Q + 1, cmp);
105     l = 1, r = 0, num = sum = 0;
106     for (int i = 1; i <= Q; ++i){
107         for (; r < a[i].r; ++r) update_right(r + 1, 1);
108         for (; r > a[i].r; --r) update_right(r, -1);
109         for (; l < a[i].l; ++l) update_left(l, -1);
110         for (; l > a[i].l; --l) update_left(l - 1, 1);
111         a[i].ans =  a[i].l == a[i].r ? 0 : sum;
112     }
113
114     sort(a + 1, a + Q + 1, cmp_id);
115     for (int i = 1; i <= Q; ++i)
116         printf("%lld\n", a[i].ans);
117     return 0;
118 }