MooFest POJ - 1990 树状数组

Every year, Farmer John's N (1 <= N <= 20,000) cows attend "MooFest",a social gathering of cows from around the world. MooFest involves a variety of events including haybale stacking, fence jumping, pin the tail on the farmer, and of course, mooing. When the cows all stand in line for a particular event, they moo so loudly that the roar is practically deafening. After participating in this event year after year, some of the cows have in fact lost a bit of their hearing. 

Each cow i has an associated "hearing" threshold v(i) (in the range 1..20,000). If a cow moos to cow i, she must use a volume of at least v(i) times the distance between the two cows in order to be heard by cow i. If two cows i and j wish to converse, they must speak at a volume level equal to the distance between them times max(v(i),v(j)). 

Suppose each of the N cows is standing in a straight line (each cow at some unique x coordinate in the range 1..20,000), and every pair of cows is carrying on a conversation using the smallest possible volume. 

Compute the sum of all the volumes produced by all N(N-1)/2 pairs of mooing cows. 

Input

* Line 1: A single integer, N 

* Lines 2..N+1: Two integers: the volume threshold and x coordinate for a cow. Line 2 represents the first cow; line 3 represents the second cow; and so on. No two cows will stand at the same location. 

Output

* Line 1: A single line with a single integer that is the sum of all the volumes of the conversing cows. 

Sample Input

4
3 1
2 5
2 6
4 3

Sample Output

57

num数组维护的是比当前位置小的牛数量
dis数组维护的是比当前位置小的牛的距离和
cnt为总距离(i - num1) * a[i].x 为大于当前牛的距离

(cnt - len) - (i - num1) * a[i].x); 

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <queue>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <set>
 7 #include <iostream>
 8 #include <map>
 9 #include <stack>
10 #include <string>
11 #include <vector>
12 #define pi acos(-1.0)
13 #define eps 1e-6
14 #define fi first
15 #define se second
16 #define lson l,m,rt<<1
17 #define rson m+1,r,rt<<1|1
18 #define bug         printf("******\n")
19 #define mem(a,b)    memset(a,b,sizeof(a))
20 #define fuck(x)     cout<<"["<<x<<"]"<<endl
21 #define f(a)        a*a
22 #define sf(n)       scanf("%d", &n)
23 #define sff(a,b)    scanf("%d %d", &a, &b)
24 #define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
25 #define pf          printf
26 #define FRE(i,a,b)  for(i = a; i <= b; i++)
27 #define FREE(i,a,b) for(i = a; i >= b; i--)
28 #define FRL(i,a,b)  for(i = a; i < b; i++)
29 #define FRLL(i,a,b) for(i = a; i > b; i--)
30 #define FIN freopen("DATA.txt","r",stdin)
31 #define lowbit(x)   x&-x
32 #pragma comment (linker,"/STACK:102400000,102400000")
33 
34 using namespace std;
35 typedef long long LL;
36 const int maxn = 2e5 + 10;
37 int num[maxn], dis[maxn], n;
38 struct node {
39     int v, x;
40 } a[maxn];
41 int cmp(node A, node B) {
42     return A.v < B.v;
43 }
44 void update(int x, int key, int *d) {
45     while(x <= 41000) {
46         d[x] += key;
47         x += lowbit(x);
48     }
49 }
50 LL sum(int x, int *d) {
51     LL ret = 0;
52     while(x > 0) {
53         ret += d[x];
54         x -= lowbit(x);
55     }
56     return  ret;
57 }
58 int main() {
59     while(scanf("%d", &n) != EOF) {
60         mem(num, 0);
61         mem(dis, 0);
62         for (int i = 0 ; i < n ; i++)
63             scanf("%d%d", &a[i].v, &a[i].x);
64         sort(a, a + n, cmp);
65         LL ans = 0, cnt = 0;
66         for (int i = 0; i < n ; i++) {
67             LL num1 = sum(a[i].x, num);
68             LL len = sum(a[i].x, dis);
69             ans += a[i].v * (num1 * a[i].x - len + (cnt - len) - (i - num1) * a[i].x);
70             cnt += a[i].x;
71             update(a[i].x, a[i].x, dis);
72             update(a[i].x, 1, num);
73         }
74         printf("%lld\n", ans);
75     }
76     return 0;
77 }