poj 1990 MooFest（树状数组）

题意：N头奶牛每头耳背程度v，坐标x。两牛交流需要音量为distance * max(v(i),v(j))，求

所有牛两两交流所需总和？

思路：对耳背进行升序，一头一头的加入牛，这样加入每一头的时候，这头牛的耳背程度

肯定是最大的。开两个树状数组，分别记录距离和牛数量。

每次分别查询左右两边的牛数量和距离和。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn = 2*1e5+5;
ll tree_dis[maxn], tree_num[maxn];
struct node
{
    ll d, v;
    bool operator < (const node &a) const
    {
        return v < a.v;
    }
}a[maxn];

int lowbit(int x) { return x&(-x); }

ll query(ll *tree, ll pos)
{
    ll sum = 0;
    while(pos)
    {
        sum += tree[pos];
        pos -= lowbit(pos);
    }
    return sum;
}

void update(ll *tree, ll pos, ll val)
{
    while(pos < maxn)
    {
        tree[pos] += val;
        pos += lowbit(pos);
    }
}

ll sum(ll *tree, ll from, ll to)
{
    return query(tree, to-1) - query(tree, from-1);
}

int main(void)
{
    int n;
    while(cin >> n)
    {
        memset(tree_num, 0, sizeof(tree_num));
        memset(tree_dis, 0, sizeof(tree_dis));
        for(int i = 0; i < n; i++)
            scanf("%lld%lld", &a[i].v, &a[i].d);
        sort(a, a+n);
        ll res = 0;
        for(int i = 0; i < n; i++)
        {
            ll v = a[i].v, d = a[i].d;
            ll l = sum(tree_num, 1, d);
            ll r = sum(tree_num, d+1, maxn);
            res += v*((l*d-sum(tree_dis, 1, d))+(sum(tree_dis, d+1, maxn)-r*d));
            update(tree_dis, a[i].d, a[i].d);
            update(tree_num, a[i].d, 1);
        }
        printf("%lld\n", res);
    }
    return 0;
}

G - MooFest

Time Limit:1000MS     Memory Limit:30000KB     64bit IO Format:%lld & %llu

Submit   Status

Description

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