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: 2000MS
Memory 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 <= i <= T, 1<= j <= M). Well, can you calculate the probability that all of the teams solve at least one problem, and at the same time the champion team solves at least N problems? > #### 输入 > The input consists of several test cases. The first line of each test case contains three integers M (0 < M <= 30), T (1 < T <= 1000) and N (0 < N <= M). Each of the following T lines contains M floating-point numbers in the range of [0,1]. In these T lines, the j-th number in the i-th line is just Pij. A test case of M = T = N = 0 indicates the end of input, and should not be processed.

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
",biger-sum);
    }
    return 0;
}

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