POJ 2151 Check the difficulty of problems 概率dp+01背包

时间：2023-03-09 19:56:42
POJ 2151 Check the difficulty of problems 概率dp+01背包

题目链接：

http://poj.org/problem?id=2151

Check the difficulty of problems

Time Limit: 2000MSMemory Limit: 65536K
#### 问题描述
> Organizing a programming contest is not an easy job. To avoid making the problems too difficult, the organizer usually expect the contest result satisfy the following two terms:
> 1. All of the teams solve at least one problem.
> 2. The champion (One of those teams that solve the most problems) solves at least a certain number of problems.
>
> Now the organizer has studied out the contest problems, and through the result of preliminary contest, the organizer can estimate the probability that a certain team can successfully solve a certain problem.
>
> Given the number of contest problems M, the number of teams T, and the number of problems N that the organizer expect the champion solve at least. We also assume that team i solves problem j with the probability Pij (1
#### 输入
> The input consists of several test cases. The first line of each test case contains three integers M (0 输出

For each test case, please output the answer in a separate line. The result should be rounded to three digits after the decimal point.

样例输入

2 2 2

0.9 0.9

1 0.9

0 0 0

样例输出

0.972

题意

acm竞赛总共有m道题，t个队伍，并且知道第i个队伍做出第j道题的概率，求保证每个队伍都至少做出一道题，并且冠军队伍至少做出n道题的概率。

题解

原问题可以转换为求：所有的队伍都至少做出一题的概率-每个队伍都做出（1~n-1)道题的总概率。

代码

#include<map>

#include<set>

#include<cmath>

#include<queue>

#include<stack>

#include<ctime>

#include<vector>

#include<cstdio>

#include<string>

#include<bitset>

#include<cstdlib>

#include<cstring>

#include<iostream>

#include<algorithm>

#include<functional>

using namespace std;

#define X first

#define Y second

#define mkp make_pair

#define lson (o<<1)

#define rson ((o<<1)|1)

#define mid (l+(r-l)/2)

#define sz() size()

#define pb(v) push_back(v)

#define all(o) (o).begin(),(o).end()

#define clr(a,v) memset(a,v,sizeof(a))

#define bug(a) cout<<#a<<" = "<<a<<endl

#define rep(i,a,b) for(int i=a;i<(b);i++)

#define scf scanf

#define prf printf

typedef __int64 LL;

typedef vector<int> VI;

typedef pair<int,int> PII;

typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;

const LL INFL=0x3f3f3f3f3f3f3f3fLL;

const double eps=1e-9;

const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=3333;

const int maxm=33;

///n个队伍，m道题，冠军至少做mm道。

int n,m,mm;

double dp[maxn][maxm];

double dp2[maxm][maxm];

double mat[maxn][maxm];

void init(){

    clr(dp,0);

}

int main() {

    while(scf("%d%d%d",&m,&n,&mm)==3&&n){

        init();

        for(int i=1;i<=n;i++){

            clr(dp2,0);

            ///01背包,dp2[j][k],求第i个队伍前j道题做出k道的概率。

            dp2[0][0]=1.0;

            for(int j=1;j<=m;j++){

                double x; scf("%lf",&x);

                for(int k=0;k<=j;k++){

                    dp2[j][k]+=dp2[j-1][k]*(1-x);

                    if(k) dp2[j][k]+=dp2[j-1][k-1]*x;

                }

            }

            ///mat[i][k]:第i个队伍m道题做出k道的概率。

            for(int k=0;k<=mm-1;k++) mat[i][k]=dp2[m][k];

        }

        ///biger:所有的队伍都至少做出一题的概率

        double biger=1.0;

        for(int i=1;i<=n;i++) biger*=(1-mat[i][0]);

        ///dp[i][j]:统计前i个队伍中第i个队伍做了j题的总概率。

        clr(dp,0);

        for(int j=1;j<=mm-1;j++) dp[1][j]=mat[1][j];

        for(int i=2;i<=n;i++){

            for(int j=1;j<=mm-1;j++){

                for(int k=1;k<=mm-1;k++){

                    dp[i][j]+=dp[i-1][k]*mat[i][j];

                }

            }

        }

        ///sum:每个队伍都做出（1~mm-1)道题的总概率

        double sum=0;

        for(int i=1;i<=mm-1;i++) sum+=dp[n][i];

        prf("%.3f\n",biger-sum);

    }

    return 0;

}

//end-----------------------------------------------------------------------