[POJ3071] Football

模拟，dp[i][j]表示第i局j获胜的概率

轻易～

#include<cstdio>
using namespace std;
#define maxn 1000 
double dp[10][maxn],p[maxn][maxn];
int n;
void getdp(int x,int y){
    int len=1<<(x-1);
    for(int i=1;i<=len;i++){
        dp[x][y+i]=dp[x][y+len+i]=0;
        for(int j=1;j<=len;j++){
            dp[x][y+i]+=p[y+i][y+len+j]*dp[x-1][y+len+j];
            dp[x][y+len+i]+=p[y+len+i][y+j]*dp[x-1][y+j];
        }
        dp[x][y+i]*=dp[x-1][y+i];
        dp[x][y+len+i]*=dp[x-1][y+len+i];
    }
}
int Max(int x,int y){ return dp[n][x]>dp[n][y]?x:y; }
int main(){
    while(1){
        scanf("%d",&n);
        if(n==-1)break;
        for(int i=1;i<=1<<n;i++)
            for(int j=1;j<=1<<n;j++)
                scanf("%lf",&p[i][j]);
        for(int i=1;i<=1<<n;i+=2)
            dp[1][i]=p[i][i+1],dp[1][i+1]=p[i+1][i];
        for(int i=2;i<=n;i++)
            for(int j=1;j<=1<<n;j+=1<<i)
                getdp(i,j-1);
        int ans=0;
        for(int i=1;i<=1<<n;i++)
            ans=Max(ans,i);
        printf("%d
",ans);
    }
}