P3531 [POI2012]LIT-Letters
很容易想到冒泡排序的交换次数为逆序对数，考虑此题如何定义“排序”，显然 $a$ 中第 $k$ 个某字母一定和 $b$ 中第 $k$ 个某字母对齐，然后直接求逆序对数。
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| #include <cstdio>
#include <vector>
using namespace std;
typedef long long LL;
template <class Type>
void read(Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
c == '-' && (flag = true);
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 3) + (x << 1) + c - '0';
flag && (x = ~x + 1);
}
template <class Type>
void write(Type x)
{
x < 0 && (putchar('-'), x = ~x + 1);
x > 9 && (write(x / 10), 0);
putchar(x % 10 + '0');
}
const int N = 1e6 + 10;
int n, w[N], tr[N];
char s[N], t[N];
void modify (int x)
{
for (; x <= n; x += x & -x)
tr[x]++;
}
int query (int x)
{
int res = 0;
for (; x; x -= x & -x)
res += tr[x];
return res;
}
int main ()
{
read(n);
scanf("%s%s", s + 1, t + 1);
static int cnt[26];
vector <int> p[26];
for (int i = 1; i <= n; i++)
p[t[i] - 'A'].push_back(i);
for (int i = 1; i <= n; i++)
w[i] = p[s[i] - 'A'][cnt[s[i] - 'A']++];
LL res = 0;
for (int i = n; i; i--)
{
res += query(w[i]);
modify(w[i]);
}
write(res);
return 0;
}
|