Codeforces 513G1 513G2 Inversions problem [概率dp]

  1 #include<iostream>
  2 #include<cstring>
  3 #include<cstdlib>
  4 #include<cstdio>
  5 #include<algorithm>
  6 #include<cmath>
  7 #include<queue>
  8 #include<map>
  9 #include<set>
 10 #include<stack>
 11 #include<string>
 12 
 13 #define N 105
 14 #define M 10005
 15 //#define mod 10000007
 16 //#define p 10000007
 17 #define mod2 1000000000
 18 #define ll long long
 19 #define ull unsigned long long
 20 #define LL long long
 21 #define eps 1e-6
 22 //#define inf 2147483647
 23 #define maxi(a,b) (a)>(b)? (a) : (b)
 24 #define mini(a,b) (a)<(b)? (a) : (b)
 25 
 26 using namespace std;
 27 
 28 int n;
 29 int k;
 30 double dp[N][N];
 31 double ans;
 32 double p;
 33 double tmp[N][N];
 34 int v[N];
 35 
 36 void ini()
 37 {
 38     ans=0;
 39     int i,j;
 40     for(i=1;i<=n;i++){
 41         scanf("%d",&v[i]);
 42     }
 43     p=1.0*n*(n+1)/2.0;
 44     memset(dp,0,sizeof(dp));
 45     for(i=1;i<=n;i++){
 46         for(j=i+1;j<=n;j++){
 47             dp[i][j]=1.0;
 48         }
 49     }
 50 }
 51 
 52 void solve()
 53 {
 54     int i,j,x,y,a,b;
 55     while(k--){
 56         memcpy(tmp,dp,sizeof(dp));
 57         memset(dp,0,sizeof(dp));
 58         for(i=1;i<=n;i++){
 59             for(j=i+1;j<=n;j++){
 60                 for(x=1;x<=n;x++){
 61                     for(y=x;y<=n;y++){
 62                         a=i;b=j;
 63                         if(x<=i && i<=y) a=x+y-a;
 64                         if(x<=j && j<=y) b=x+y-b;
 65                         if(a>b) swap(a,b);
 66                         if(x<=i && j<=y){
 67                             dp[a][b]+=(1.0-tmp[i][j])/p;
 68                         }
 69                         else{
 70                             dp[a][b]+=1.0*tmp[i][j]/p;
 71                         }
 72                     }
 73                 }
 74             }
 75         }
 76     }
 77 }
 78 
 79 void out()
 80 {
 81     int i,j;
 82     for(i=1;i<=n;i++){
 83         for(j=i+1;j<=n;j++){
 84             if(v[i]>v[j]){
 85                 ans+=dp[i][j];
 86             }
 87             else{
 88                 ans+=1.0-dp[i][j];
 89             }
 90         }
 91     }
 92     printf("%.10f\n",ans);
 93 }
 94 
 95 int main()
 96 {
 97     //freopen("data.in","r",stdin);
 98     //freopen("data.out","w",stdout);
 99     //scanf("%d",&T);
100     //for(int ccnt=1;ccnt<=T;ccnt++)
101     //while(T--)
102     //scanf("%d%d",&n,&m);
103     while(scanf("%d%d",&n,&k)!=EOF)
104     {
105         ini();
106         solve();
107         out();
108     }
109     return 0;
110 }