POJ 2151 Check the difficulty of problems(概率)
2013-05-03 20:42
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题目链接:http://poj.org/problem?id=2151
题意:K个队伍参加ACM比赛,有n道题。Pij表示第i个队伍AC第j道题的概率。有所有队伍都通过至少一题且过题最多的队伍过题数大于等于m的概率。
思路:首先对于每个队伍计算f[i][j]表示前i道题AC了j道 的概率,然后用这个数组更新dp[i][j][k],前i个队伍,过题数最少为j最多为k的概率。
题意:K个队伍参加ACM比赛,有n道题。Pij表示第i个队伍AC第j道题的概率。有所有队伍都通过至少一题且过题最多的队伍过题数大于等于m的概率。
思路:首先对于每个队伍计算f[i][j]表示前i道题AC了j道 的概率,然后用这个数组更新dp[i][j][k],前i个队伍,过题数最少为j最多为k的概率。
int n,m,K; double f[35][35],dp[2][35][35],p[35]; int main() { Rush(n,K,m) { if(!n&&!m&&!K) break; int i,j,k,t,pre=0,cur=1; FOR1(k,K) { FOR1(i,n) RD(p[i]); clr(f,0); f[0][0]=1; FOR1(i,n) FOR0(j,n+1) { f[i][j]=f[i-1][j]*(1-p[i]); if(j>0) f[i][j]+=f[i-1][j-1]*p[i]; } clr(dp[cur],0); if(k==1) { FOR0(i,n+1) dp[cur][i][i]=f [i]; } else { FOR0(i,n+1) FOR0(j,n+1) for(t=j;t<=n;t++) { dp[cur][min(i,j)][max(t,i)]+=dp[pre][j][t]*f [i]; } } pre^=1; cur^=1; } double ans=0; for(i=1;i<=n;i++) for(j=m;j<=n;j++) ans+=dp[pre][i][j]; PR(ans); } return 0; }
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