MemSQL Start[c]UP 3.0

从第上往下做dfs
把比赛的关系图看成一个二叉树
对于[o][i] 表示在编号为o的比赛中i获胜了
算出所有可能性 把概率加起来
乘上新加的分数就是新加的分数的期望

#include <iostream>#include <algorithm>#include <sstream>#include <string>#include <queue>#include <cstdio>#include <map>#include <set>#include <utility>#include <stack>#include <cstring>#include <cmath>#include <vector>#include <ctime>#include <bitset>using namespace std;#define pb push_back#define sd(n) scanf("%d",&n)#define sdd(n,m) scanf("%d%d",&n,&m)#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)#define sld(n) scanf("%lld",&n)#define sldd(n,m) scanf("%lld%lld",&n,&m)#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)#define sf(n) scanf("%lf",&n)#define sff(n,m) scanf("%lf%lf",&n,&m)#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)#define ss(str) scanf("%s",str)#define ans() printf("%d",ans)#define ansn() printf("%d\n",ans)#define anss() printf("%d ",ans)#define lans() printf("%lld",ans)#define lanss() printf("%lld ",ans)#define lansn() printf("%lld\n",ans)#define fansn() printf("%.10f\n",ans)#define r0(i,n) for(int i=0;i<(n);++i)#define r1(i,e) for(int i=1;i<=e;++i)#define rn(i,e) for(int i=e;i>=1;--i)#define rsz(i,v) for(int i=0;i<(int)v.size();++i)#define szz(x) ((int)x.size())#define mst(abc,bca) memset(abc,bca,sizeof abc)#define lowbit(a) (a&(-a))#define all(a) a.begin(),a.end()#define pii pair<int,int>#define pli pair<ll,int>#define mp make_pair#define lrt rt<<1#define rrt rt<<1|1#define X first#define Y second#define PI (acos(-1.0))#define sqr(a) ((a)*(a))typedef long long ll;typedef unsigned long long ull;const ll mod = 1000000000+7;const double eps=1e-9;const int inf=0x3f3f3f3f;const ll infl = 10000000000000000;const int maxn=  1000+10;const int maxm = 1000+10;//Pretests passedint in(int &ret){    char c;    int sgn ;    if(c=getchar(),c==EOF)return -1;    while(c!='-'&&(c<'0'||c>'9'))c=getchar();    sgn = (c=='-')?-1:1;    ret = (c=='-')?0:(c-'0');    while(c=getchar(),c>='0'&&c<='9')ret = ret*10+(c-'0');    ret *=sgn;    return 1;}double win[maxn][maxn];double e[maxn][maxn];double p[maxn][maxn];void dfs(int o,int l,int r){    if(l==r)    {        win[o][l] = 1;        e[o][l] = 0;        return ;    }    int mid = (l+r)>>1;    dfs(o<<1  , l,   mid);    dfs(o<<1|1, mid+1 ,r);    for(int i = l;i<=mid;++i)    {        for(int j = mid+1;j<=r;++j)        {            win[o][i] += win[o<<1][i]*win[o<<1|1][j]*p[i][j];        }    }    for(int i = l;i<=mid;++i)    {        for(int j = mid+1;j<=r;++j)        {            e[o][i] = max(e[o][i],win[o][i]*(r-l+1)/2+e[o<<1][i]+e[o<<1|1][j]);        }    }    for(int i = mid+1;i<=r;++i)    {        for(int j = l;j<=mid;++j)        {            win[o][i] += win[o<<1|1][i]*win[o<<1][j]*p[i][j];        }    }    for(int i = mid+1;i<=r;++i)    {        for(int j = l;j<=mid;++j)        {            e[o][i] = max(e[o][i],win[o][i]*(r-l+1)/2+e[o<<1|1][i]+e[o<<1][j]);        }    }}int main(){#ifdef LOCAL    freopen("input.txt","r",stdin);//    freopen("output.txt","w",stdout);#endif // LOCAL    int n;    sd(n);    n = 1<<n;    r1(i,n)r1(j,n)    {        int x;        sd(x);        p[i][j] = 0.01*x;    }    dfs(1,1,n);    double ans = 0;    r1(i,n)ans = max(ans,e[1][i]);    printf("%.10f\n",ans);    return 0;}