MooFest POJ

题意：n 头牛，每两头牛之间产生的价值是 max(a[i].v , a[j].v ) * |a[i].x - a[j].x|，问产生的总费用是多少

题解：先将每头牛按照 v 的大小从小到大排序，那么就可以去掉一个绝对值并且只用当前的v的值去乘，之后用两个树状数组维护，一个树状数组维护比a[i].x小的牛有几头，另一个维护比a[i].x小的 x 的总和，之后 a[i].v * (sum1(a[i].x) * a[i].x - sum2(a[i].x)) 就是比它 x 小的费用总和，a[i].v * ((sum2[20000]  - sum2(a[i].x)) - sum1(20000) - sum1(a[i].x)) * a[i].x) 就是后面的总费用

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<list>
#include<math.h>
#include<vector>
#include<stack>
#include<string>
#include<cstring>
#include<string.h>
using namespace std;
typedef long long LL;
#define ll LL
#define bug printf("*********
")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
const int MAXN = 2e4 + 10;
const double eps = 1e-8;
const LL mod = 1000000007;

LL sum1[MAXN],sum2[MAXN];
struct node {
    int x,v;
}a[MAXN];
int lowbit(int x) {
    return x & -x;
}

void update1(int x,int y) {
    for(int i = x; i <= MAXN; i += lowbit(i))
        sum1[i] += y;
}
void update2(int x,int y) {
    for(int i = x; i <= MAXN; i += lowbit(i))
        sum2[i] += y;
}

LL getsum1(int x) {
    LL ans = 0;
    for(int i = x; i; i -= lowbit(i))
        ans += sum1[i];
    return ans;
}

LL getsum2(int x) {
    LL ans = 0;
    for(int i = x; i; i -= lowbit(i))
        ans += sum2[i];
    return ans;
}

void init() {
    memset(sum1,0,sizeof sum1);
    memset(sum2,0,sizeof sum2);
}
bool cmp(node x,node y) {
    return x.v < y.v;
}
int main()
{
    init();
    int n;
    int M = 2e4;
    scanf("%d",&n);
    for(int i = 1; i <= n; i++)
        scanf("%d %d",&a[i].v, &a[i].x);
    sort(a + 1, a + 1 + n,cmp);

    LL ans = 0;
    for(int i = 1; i <= n; i++) {
//        debug(getsum1(a[i].x));
//        debug(getsum2(a[i].x));
//        debug(getsum1(M) - getsum1(a[i].x));
//        debug(getsum2(M) - getsum2(a[i].x));
        ans += a[i].v * (getsum1(a[i].x) * a[i].x - getsum2(a[i].x));
//        debug(ans);
        ans += a[i].v * ((getsum2(M) - getsum2(a[i].x)) - (getsum1(M) - getsum1(a[i].x)) * a[i].x);
        update1(a[i].x,1);
        update2(a[i].x,a[i].x);
//        debug(ans);
    }
    printf("%lld
",ans);
}