【HDOJ】1394 Minimum Inversion Number

逆序数的性质。
1. 暴力解

#include <stdio.h>

#define MAXNUM 5005

int a[MAXNUM];

int main() {
    int n;
    int i, j, sum, min;

    while (scanf("%d", &n) != EOF) {
        for (i=0; i<n; ++i)
            scanf("%d", &a[i]);
        sum = 0;
        for (i=0; i<n; ++i) {
            for (j=i+1; j<n; ++j)
                if (a[j] < a[i])
                    ++sum;
        }
        min = sum;
        for (i=0; i<n; ++i) {
            sum += n - 1 - a[i]*2;
            if (sum < min)
                min = sum;
            //printf("%d
", sum);
        }
        printf("%d
", min);
    }

    return 0;
}

2. 线段树。需要理解如何求逆序数。

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1

const int maxn = 5005;

int nums[maxn<<2];
int a[maxn];

void PushUP(int rt) {
    nums[rt] = nums[rt<<1] + nums[rt<<1|1];
}

void build(int l, int r, int rt) {
    nums[rt] = 0;
    if (l == r) return;
    int m = (l+r)>>1;
    build(lson);
    build(rson);
}

void update(int p, int l, int r, int rt) {
    if (l == r) {
        nums[rt] = 1;
        return ;
    }
    int m = (l+r)>>1;
    if (p <= m)
        update(p, lson);
    else
        update(p, rson);
    PushUP(rt);
}

int query(int ll, int rr, int l, int r, int rt) {
    if (ll<=l && rr>=r) {
        return nums[rt];
    }
    int m = (l+r)>>1;
    int ret = 0;
    if (ll <= m)
        ret += query(ll, rr, lson);
    if (rr > m)
        ret += query(ll, rr, rson);

    return ret;
}

int main() {
    int n;
    int i, sum, min;

    while (scanf("%d", &n) != EOF) {
        build(0, n-1, 1);
        sum = 0;
        for (i=0; i<n; ++i) {
            scanf("%d", &a[i]);
            sum += query(a[i], n-1, 0, n-1, 1);
            update(a[i], 0, n-1, 1);
            //printf("%d: sum=%d
", i, sum);
        }
        min = sum;
        for (i=0; i<n; ++i) {
            sum += n - 1 - a[i]*2;
            if (sum < min)
                min = sum;
        }
        printf("%d
", min);
    }

    return 0;
}