HDU 4911 （树状数组）

题目连接：点击这里

题意：给出一个序列， 交换某些相邻的数， 求剩下逆序对最少的个数。

显然一次交换可以减少一个逆序对， 所以离散化一下树状数组一下求出最大的逆序对减去交换次数即可。trick是判断交换次数比逆序对大的情况。

#include <bits/stdc++.h>
using namespace std;
#define maxn 100005

int n, k;
map <int, int> gg;
vector <int> num;
struct node {
int num, id;
}a[maxn];
int c[maxn];

int lowbit (int x) {
return x&(-x);
}

void add (int pos) {
for (int i = pos; i <= n; i += lowbit (i)) {
c[i]++;
}
}

int sum (int pos) {
int ans = 0;
for (int i = pos; i >= 1; i -= lowbit (i)) {
ans += c[i];
}
return ans;
}

bool cmp1 (const node &a, const node &b) {
return a.num > b.num || (a.num == b.num && a.id > b.id);
}

void lisanhua () {
gg.clear ();
sort (num.begin (), num.end ());
int cnt = 0;
for (int i = 0; i < n; i++) {
if (!i || num[i] != num[i-1]) gg[num[i]] = ++cnt;
}
for (int i = 1; i <= n; i++) a[i].num = gg[a[i].num];
}

int main () {
while (scanf ("%d%d", &n, &k) == 2) {
num.clear ();
for (int i = 1; i <= n; i++) {
scanf ("%d", &a[i].num);
num.push_back (a[i].num);
a[i].id = i;
}
lisanhua ();
sort (a+1, a+1+n, cmp1);
memset (c, 0, sizeof c);
long long ans = 0;
for (int i = 1; i <= n; i++) {
ans += sum (a[i].id);
add (a[i].id);
}
if (ans-k <= 0) {
printf ("0\n");
}
else
printf ("%lld\n", ans-k);
}
return 0;
}