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); 

#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <algorithm>
#include <set>
#include <iostream>
#include <map>
#include <stack>
#include <string>
#include <vector>
#define pi acos(-1.0)
#define eps 1e-6
#define fi first
#define se second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define bug         printf("******\n")
#define mem(a,b)    memset(a,b,sizeof(a))
#define fuck(x)     cout<<"["<<x<<"]"<<endl
#define f(a)        a*a
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define FRE(i,a,b)  for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b)  for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define FIN freopen("DATA.txt","r",stdin)
#define lowbit(x)   x&-x
#pragma comment (linker,"/STACK:102400000,102400000")

using namespace std;
typedef long long LL;
const int maxn = 2e5 + 10;
int num[maxn], dis[maxn], n;
struct node {
    int v, x;
} a[maxn];
int cmp(node A, node B) {
    return A.v < B.v;
}
void update(int x, int key, int *d) {
    while(x <= 41000) {
        d[x] += key;
        x += lowbit(x);
    }
}
LL sum(int x, int *d) {
    LL ret = 0;
    while(x > 0) {
        ret += d[x];
        x -= lowbit(x);
    }
    return  ret;
}
int main() {
    while(scanf("%d", &n) != EOF) {
        mem(num, 0);
        mem(dis, 0);
        for (int i = 0 ; i < n ; i++)
            scanf("%d%d", &a[i].v, &a[i].x);
        sort(a, a + n, cmp);
        LL ans = 0, cnt = 0;
        for (int i = 0; i < n ; i++) {
            LL num1 = sum(a[i].x, num);
            LL len = sum(a[i].x, dis);
            ans += a[i].v * (num1 * a[i].x - len + (cnt - len) - (i - num1) * a[i].x);
            cnt += a[i].x;
            update(a[i].x, a[i].x, dis);
            update(a[i].x, 1, num);
        }
        printf("%lld\n", ans);
    }
    return 0;
}