HDU 2242(概率dp 分情况统计)

来源：互联网发布：淘宝gucci天猫旗舰店编辑：程序博客网时间：2024/06/05 06:34

本题目的意思:

给定M(M<=30 )道题目要求解，有T(T <= 1000 )只队伍参赛，并给定每个队伍AC每个题的概率，求每个队伍至少AC 1道题目，并且冠军至少AC N到题目的概率；

分析：

首先对每只队伍，求出至少AC一题，但AC题数比N小的概率p1，和AC至少N题的概率p;

这样就可以用简单背包解决该问题，

定义d[ i ][ j ] 代表前面i-1个已经决策，当前存不存在已经AC大于等于N的队伍（j=1 代表有，j = 0 代表没有）

那么d[ i ][ j ] = d[ i+1 ][ 1 ] * p2[ i ] + d[ i+1 ] [ j ] * p1[i] ;

//#pragma comment(linker, "/STACK:1024000000,1024000000")#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <vector>#include <set>#include <map>#include <string>#include <list>#include <cstdlib>#include <queue>#include <stack>#include <cmath>#define ALL(a) a.begin(), a.end()#define clr(a, x) memset(a, x, sizeof a)#define fst first#define snd second#define pb push_back#define lowbit(x) (x&(-x))#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define rep1(i,x,y) for(int i=x;i<=y;i++)#define rep(i,n) for(int i=0;i<(int)n;i++)using namespace std;const double eps = 1e-10;typedef long long LL;typedef long long ll;typedef pair<int, int> pii;const int inf =0x3f3f3f3f;const int T = 1010;const int M = 35;int lim,m,t,n,now;double d[M][M],p1[T],p2[T],p[T][M],r[T][2];bool vis[M][M] , vv[T][2];double dp2(int i,int j){   if(vis[i][j]) return d[i][j];   vis[i][j]=true;   if(i == m+1) return d[i][j] = (j>=n);   d[i][j]=dp2(i+1,j)*(1-p[now][i]);   d[i][j]+=dp2(i+1,j+1)*p[now][i];   return d[i][j];}double dp(int i,int j){   if(vv[i][j]) return r[i][j];   vv[i][j] = true;   if(i == t+1) return r[i][j] = j;   r[i][j] = dp(i+1,j)*p1[i] + dp(i+1,1)*p2[i];   return r[i][j];}int main(){   while(scanf("%d %d %d",&m,&t,&n)==3 && m){      rep1(i,1,t) rep1(j,1,m)         scanf("%lf",&p[i][j]);      rep1(i,1,t){          now = i;          memset(vis,false,sizeof(vis));          p2[i] = dp2(1,0);          double te = 1; rep1(j,1,m) te *= (1-p[i][j]);          p1[i] = 1 - p2[i] - te;      }      memset(vv,false,sizeof(vv));      printf("%.3lf\n",dp(1,0));   }   return 0;}

0 0