概率期望总结

例题:

1.POJ 3744
详解:
2.POJ - 3071 Football
代码:

#include <cstdio>
#include <cstring>
using namespace std;
const int N=(1<<7)+5;
double p[N][N],win[N][10];
int main()
{
    int n;
    while(scanf("%d",&n),n!=-1)
    {
        int m=1<<n;
        double maxn=0;
        int num=1;
        for(int i=1;i<=m;i++)
        {
            for(int j=1;j<=m;j++)
                scanf("%lf",&p[i][j]);
        }
        memset(win,0,sizeof(win));
        for(int i=1;i<=m;i++)
            win[i][0]=1;
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {//cout<<"j="<<j<<" : ";
                int len=1<<i;
                int s=j/len;
                if(j%len==0)
                    s--;
                s*=len;//cout<<s+1<<"->"<<s+len<<" ";
                int l=(j>(s+len/2)?(s+1):(s+len/2+1));
                int r=(j>s+len/2?s+len/2:s+len);
                //cout<<"l= "<<l<<" r="<<r<<" ";
                for(int k=l;k<=r;k++)//第i轮第j个选手需要比较的区间
                {
                    if(k==j)
                        continue;
                    win[j][i]+=win[j][i-1]*win[k][i-1]*p[j][k];
                }
                //cout<<win[j][i]<<endl;
                if(i==n&&win[j][i]>maxn)
                {
                    maxn=win[j][i];
                    num=j;
                }
            }
        }//cout<<maxn<<endl;
        printf("%d\n",num);
    }
    return 0;
}

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转载自www.cnblogs.com/1024-xzx/p/12377965.html