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;} |
原文：http://www.cnblogs.com/rolight/p/3528467.html