2011 Heilongjiang collegiate programming contest - Problem H - Car race game

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比赛解题报告同时被 2 个专栏收录

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树状数组

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本文深入解析了两种用于计算数列逆序数的复杂度为O(nlogn)的算法：树状数组与归并排序。通过离散化处理与优化技巧，实现快速准确地统计逆序数，旨在提升数据结构与算法领域的实践效率。

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求数列逆序数复杂度O(nlogn)的两种算法

1. 树状数组，对于大数据需要进行离散化处理

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <vector>
#include <stack>
#define INF (int)(1e9)
#define maxn 1000010
#define Lson l, mid, rt << 1
#define Rson mid + 1, r, rt << 1 | 1
using namespace std;
typedef long long ll;
struct node {
    int x, v;
    bool operator < (const node &g) const {
        if (x != g.x) return x < g.x;
        return v > g.v;
    }
}no[maxn];
int num[maxn << 1], n;
int lowbit(int x) {return x&(-x);}
void update(int x, int v) {
    for (; x <= maxn; x += lowbit(x)) {
        num[x] += v;
    }
}
ll query(int x) {
    ll ret = 0;
    for (; x > 0; x -= lowbit(x)) {
        ret += ll(num[x]);
    }
    return ret;
}
int main() {
    while (scanf("%d", &n) != EOF) {
        memset(num, 0, sizeof(num));
        for (int i = 1; i <= n; ++ i) {
            scanf("%d%d", &no[i].x, &no[i].v);
        }
        sort(no+1, no+n+1);
        ll res = 0;
        for (int i = 1; i <= n; ++ i) {
            update(no[i].v,1);
            res += (i - query(no[i].v));
        }
        printf("%lld\n", res);

    }
}

2. 归并排序同时求出逆序数

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <vector>
#include <stack>
#define INF (int)(1e9)
#define maxn 1000010
#define Lson l, mid, rt << 1
#define Rson mid + 1, r, rt << 1 | 1
using namespace std;
typedef long long ll;
struct P {
    int a,b;
} s[1100005],t[1100005];
bool cmp (P a, P b) {
    if (a.a != b.a) return a.a < b.a;
    return a.b > b.b;
}
ll ans;
void merge_sort(P* A, int x, int y, P *T) {
    if (y - x > 1) {
        int m = (y + x) >> 1;
        int p = x, q = m, i = x;
        merge_sort(A, x, m, T);
        merge_sort(A, m, y, T);
        while (p < m || q < y) {
            if(q >= y || (p < m && A[p].b <= A[q].b)) {
                T[i++] = A[p++];
            } else {
                T[i++] = A[q++];
                ans += ll(m-p);
            }
        }
        for (i = x; i < y; i ++)
            A[i] = T[i];
    }
}
int main() {
    int n;
    while (cin >> n) {
        memset(s, 0, sizeof(s));
        memset(t, 0, sizeof(t));
        for(int i = 0; i < n; ++ i) {
            scanf("%d%d", &s[i].a, &s[i].b);
        }
        sort(s, s+n, cmp);
        ans = 0;
        merge_sort(s, 0, n, t);
        cout << ans << endl;
    }
    return 0;
}