大体思路就是枚举裁判位置，找逆序对数。
练习一下用树状数组来寻找逆序对
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#include <cstdio>#include <cstring>#include <iostream>#define lowbit(x) ((x)&(-x))using
namespace
std;
const
int
maxi = 100005;
const
int
maxn = 20001;
int
C[maxi],num[maxn],n;
int
l[maxn],r[maxn];
inline
void
add(int
x,int
d) {
while(x <= maxi) {
C[x] += d; x += lowbit(x);
}
}inline
int
sum(int
x) {
int
ans = 0;
while(x > 0) {
ans += C[x]; x -= lowbit(x);
}
return
ans;
}int
main() {
int
T; scanf("%d",&T);
while(T--) {
long
long
ans = 0;
scanf("%d",&n);
memset(C,0,sizeof(C));
for(int
i = 1;i <= n;i++) scanf("%d",num + i);
for(int
i = 1;i <= n;i++) {
add(num[i],1);
l[i] = sum(num[i] - 1);
}
memset(C,0,sizeof(C));
for(int
i = n;i >= 1;i--) {
add(num[i],1);
r[i] = sum(num[i] - 1);
}
for(int
i = 2;i < n;i++) {
ans += l[i] * (n - i - r[i]) + r[i] * (i - 1 - l[i]);
}
cout << ans << endl;
}
return
0;
} |