題意：
給你一個長度爲n(<=100)的序列T，S爲T的任意子序列，r(S)表示子序列S(不連續)中的逆序對數，l(S) 表示S的長度，求出 r(S) / l(S) 的最大值。
思路：
將r(S)看成邊，l(S)看成點，問題轉化爲求 E / V 的最大值。經典的最大密度子圖問題。
利用類似0/1分數規劃的思想，二分答案，設爲mid，則有E / V=mid 即E=V*mid。
即使E-V*mid趨近於0。
問題再轉化爲求最大權閉合圖。最大權閉合圖參考：https://blog.csdn.net/LSD20164388/article/details/79224422
設源點爲st，匯點爲ed，建邊：
1、對於每個逆序對標號i=1~m，依次建邊( st , i , 1.0 )
2、對於每個逆序對標號i=1~m，對應原數組下標爲(x,y)，依次建邊 ( i , x+m , inf ) ， ( i , y+m , inf )
3、對於每個點i=1~n，依次建邊( i+m , ed , mid )
使用sap模板的代碼：
#include<bits/stdc++.h>
#define ll long long
#define inf 0x3f3f3f3f
#define rep(i,a,b) for(register int i=(a);i<=(b);i++)
#define dep(i,a,b) for(register int i=(a);i>=(b);i--)
using namespace std;
#define maxn 200010
#define clear(A, X) memset (A, X, sizeof A)
#define copy(A, B) memcpy (A, B, sizeof A)
using namespace std;
const double eps=1e-9;
template <typename T>
inline void read(T &X)
{
X=0;int w=0; char ch=0;
while(!isdigit(ch)) {w|=ch=='-';ch=getchar();}
while(isdigit(ch)) X=(X<<3)+(X<<1)+(ch^48),ch=getchar();
if(w) X=-X;
}
pair<int,int>ne[maxn];
const int MAXN = 20010; //點數的最大值
const int MAXM = 40010; //邊數的最大值
const int INF = 0x3f3f3f3f;
struct Edge {
int to, next;double cap, flow;
} edge[MAXM]; //注意是MAXM
int tol;
int head[MAXN];
int gap[MAXN], dep[MAXN], pre[MAXN], cur[MAXN];
int n,m;
//加邊，單向圖三個參數，雙向圖四個參數
void addedge(int u, int v, double w, double rw = 0.0) {
edge[tol].to = v;
edge[tol].cap = w;
edge[tol].next = head[u];
edge[tol].flow = 0;
head[u] = tol++;
edge[tol].to = u;
edge[tol].cap = rw;
edge[tol].next = head[v];
edge[tol].flow = 0;
head[v] = tol++;
}
//輸入參數：起點、終點、點的總數
//點的編號沒有影響，只要輸入點的總數
double sap(int start, int end, int N) {
memset(gap, 0, sizeof(gap));
memset(dep, 0, sizeof(dep));
memcpy(cur, head, sizeof(head));
int u = start;
pre[u] = -1;
gap[0] = N;
double ans = 0.0;
while (dep[start] < N) {
if (u == end) {
double Min = INF;
for (int i = pre[u]; i != -1; i = pre[edge[i ^ 1].to])
if (Min > edge[i].cap - edge[i].flow)
Min = edge[i].cap - edge[i].flow;
for (int i = pre[u]; i != -1; i = pre[edge[i ^ 1].to]) {
edge[i].flow += Min;
edge[i ^ 1].flow -= Min;
}
u = start;
ans += Min;
continue;
}
bool flag = false;
int v;
for (int i = cur[u]; i != -1; i = edge[i].next) {
v = edge[i].to;
if (edge[i].cap - edge[i].flow && dep[v] + 1 == dep[u]) {
flag = true;
cur[u] = pre[v] = i;
break;
}
}
if (flag) {
u = v;
continue;
}
int Min = N;
for (int i = head[u]; i != -1; i = edge[i].next)
if (edge[i].cap - edge[i].flow && dep[edge[i].to] < Min) {
Min = dep[edge[i].to];
cur[u] = i;
}
gap[dep[u]]--;
if (!gap[dep[u]])
return ans;
dep[u] = Min + 1;
gap[dep[u]]++;
if (u != start)
u = edge[pre[u] ^ 1].to;
}
return ans;
}
int ru[MAXN],fa[MAXN];
void init () {//初始化
tol = 0;
clear (head, -1);
clear (fa, -1);
clear (ru, 0);
}
double build(int st,int ed,double mid){
init();
rep(i,1,m) addedge(st,i,1.0);
rep(i,1,m) {
addedge(i,ne[i].first+m,(double)inf);
addedge(i,ne[i].second+m,(double)inf);
}
rep(i,1,n) addedge(i+m,ed,mid);
return m-sap(n+m+1,n+m+2,n+m+2);
}
int a[maxn];
int main(){
int i,j,k,l,x,y,z,cas=0,t;
read(t);
while (t--){
read(n);m=0;
for (i=1;i<=n;i++) read(a[i]);
for (i=1;i<=n;i++)
for (j=i+1;j<=n;j++){
if(a[i]>a[j]){
ne[++m].first=i;
ne[m].second=j;
}
}
double l=0,r=m;
while((r-l)>eps){
double mid=(l+r)/2;
//cout<<mid<<" "<<l<<" "<<r<<endl;
if(build(n+m+1,n+m+2,mid)>0){
l=mid;
}
else r=mid;
}
printf("Case #%d: %.12lf\n",++cas,l);
}
return 0;
}