hdu 4911 Inversion(归并)

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题目链接：hdu 4911 Inversion

题目大意：给定一个序列，有k次机会交换相邻两个位置的数，问说最后序列的逆序对数最少为多少。

解题思路：每交换一次一定可以减少一个逆序对，所以问题转换成如何求逆序对数。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
typedef long long ll;
const int maxn = 1e5+5;

ll k;
int n, s[maxn], t[maxn];

ll merge_sort (int l, int r, int* a, int* b) {

    if (l == r)
        return 0;

    int mid = (l + r) / 2;
    ll ret = merge_sort(l, mid, a, b) + merge_sort(mid+1, r, a, b);
    int mvl = l, mvr = mid+1, mv = l;
    while (mvl <= mid || mvr <= r) {
        if (mvr > r || (mvl <= mid && a[mvl] <= a[mvr])) {
            b[mv++] = a[mvl++];
        } else {
            ret += mid - mvl + 1;
            b[mv++] = a[mvr++];
        }
    }

    for (int i = l; i <= r; i++)
        a[i] = b[i];
    return ret;
}

int main () {
    while (scanf("%d%I64d", &n, &k) == 2) {
        for (int i = 1; i <= n; i++)
            scanf("%d", &s[i]);
        ll ans = merge_sort(1, n, s, t);
        printf("%I64d\n", max(ans - k, 0LL));
    }
    return 0;
}

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hdu 4911 Inversion(归并)

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原文地址：http://blog.csdn.net/keshuai19940722/article/details/38391765