Codeforces 1311F Moving Points

题目链接
根据题意，d是两个点的最短距离，分析知，假设\(x_i\)<\(x_j\), 若\(v_i\)>\(v_j\)，那么d(i,j)一定为0，因为i一定能追上j，否则，d(i,j)就为其初始距离
那我们就转化问题为一个二维偏序问题，求\(x_i\)<\(x_j\)且\(v_i\)<\(v_j\)，满足这个条件的每个点对之间的距离
很容易想到定一序，另一序用树状数组维护的统计法，假设现在是\(x_i\),满足上述条件有k个，那么，对\(x_i\)的统计答案为：\(\sum_{j=i-k}^{i-1}(x_i-x_j)\),拆开，就是\(k*x_i-\sum_{j=i-k}^{i-1}x_j\)
那我们可以维护2个树状数组，分别维护上述的2个数，满足条件的个数与满足条件的这些数的x的和即可

#include<bits/stdc++.h>
using namespace std;
#define lowbit(x) ((x)&(-x))
typedef long long LL;
typedef pair<int,int> pii;

const int maxn = 2e5+5;

struct Node {
    int x, v;
} Nodes[maxn];

int all_x[maxn];
LL C1[maxn], C2[maxn];

void add(LL val, int pos, int n) {
    for(; pos <= n; pos += lowbit(pos))
        C1[pos] += val, C2[pos]++;
}

pair<LL,LL> getsum(int pos) {
    LL ret1 = 0, ret2 = 0;
    for(;pos;pos-=lowbit(pos))
        ret1 += C1[pos], ret2 += C2[pos];
    return make_pair(ret1,ret2);
}

void run_case() {
    int n; 
    cin >> n;
    for(int i = 1; i <= n; ++i) {
        cin >> Nodes[i].x;
        all_x[i] = Nodes[i].x;
    }
    for(int i = 1; i <= n; ++i) {
        cin >> Nodes[i].v;
    }
    sort(all_x+1, all_x+1+n);
    int len = unique(all_x+1, all_x+1+n) - all_x - 1;
    for(int i = 1; i <= n; ++i) {
        Nodes[i].x = lower_bound(all_x+1, all_x+1+len, Nodes[i].x) - all_x;
    }
    sort(Nodes+1, Nodes+1+n, [&](Node &a, Node&b) {
        return a.v < b.v || (a.v == b.v && a.x < b.x);
    });
    LL ans = 0;
    for(int i = 1; i <= n; ++i) {
        add(all_x[Nodes[i].x], Nodes[i].x, n);
        auto now = getsum(Nodes[i].x-1);
        ans += 1LL*now.second*all_x[Nodes[i].x] - now.first;
    }
    cout << ans;
}

int main() {
    ios::sync_with_stdio(false), cin.tie(0);
    cout.flags(ios::fixed);cout.precision(10);
    //int t; cin >> t;
    //while(t--)
    run_case();
    cout.flush();
    return 0;
}

Codeforces 1311F Moving Points

原文：https://www.cnblogs.com/GRedComeT/p/12361988.html