hdu5045 Contest ---- 状态DP

原题链接： http://acm.hdu.edu.cn/showproblem.php?pid=5045

一：原题内容

Problem Description

In the ACM International Collegiate Programming Contest, each team consist of three students. And the teams are given 5 hours to solve between 8 and 12 programming problems.

On Mars, there is programming contest, too. Each team consist of N students. The teams are given M hours to solve M programming problems. Each team can use only one computer, but they can’t cooperate to solve a problem. At the beginning of the ith hour, they will get the ith programming problem. They must choose a student to solve this problem and others go out to have a rest. The chosen student will spend an hour time to program this problem. At the end of this hour, he must submit his program. This program is then run on test data and can’t modify any more.

Now, you have to help a team to find a strategy to maximize the expected number of correctly solved problems.

For each problem, each student has a certain probability that correct solve. If the i^th student solve the j^th problem, the probability of correct solve is P_ij .

At any time, the different between any two students’ programming time is not more than 1 hour. For example, if there are 3 students and there are 5 problems. The strategy {1,2,3,1,2}, {1,3,2,2,3} or {2,1,3,3,1} are all legal. But {1,1,3,2,3},{3,1,3,1,2} and {1,2,3,1,1} are all illegal.

You should find a strategy to maximize the expected number of correctly solved problems, if you have know all probability

 

Input

The first line of the input is T (1 ≤ T ≤ 20), which stands for the number of test cases you need to solve.

The first line of each case contains two integers N ,M (1 ≤ N ≤ 10,1 ≤ M ≤ 1000),denoting the number of students and programming problem, respectively.

The next N lines, each lines contains M real numbers between 0 and 1 , the j^th number in the i^th line is P_ij .

Output

For each test case, print a line “Case #t: ”(without quotes, t means the index of the test case) at the beginning. Then a single real number means the maximal expected number of correctly solved problems if this team follow the best strategy, to five digits after the decimal point. Look at the output for sample input for details.

Sample Input

12 30.6 0.3 0.40.3 0.7 0.9

 

Sample Output

Case #1: 2.20000

二：分析理解

有n个人做m道题目，给出每个人对每道题答对的概率，求最后答出所有题目概率的最大值。要求每两个人之间答题的数目不能超过1

dp[i][j]表示前i道题j状态时的最大值，注意j是n个人状态，位运算。

三：AC代码

#define _CRT_SECURE_NO_DEPRECATE #include<iostream>#include<vector>#include<cstring>#include<queue>#include<stack>#include<algorithm>#include<cmath>#define INF 99999999#define eps 0.0001using namespace std;int t;int n, m;double s[20][2005];double dp[1005][1050];int main(){scanf("%d", &t);for (int cas = 1; cas <= t; cas++){scanf("%d%d", &n, &m);for (int i = 1; i <= n; i++)for (int j = 1; j <= m; j++)scanf("%lf", &s[i][j]);for (int i = 0; i < 1005; i++)for (int j = 0; j < 1050; j++)dp[i][j] = -1.0;dp[0][0] = 0.0;for (int i = 1; i <= m; i++){for (int j = 0; j < (1 << n); j++){if (dp[i - 1][j] < 0)continue;for (int k = 0; k < n; k++){if (!(j&(1 << k)))//k+1没做{int temp = j | (1 << k);if (temp == (1 << n) - 1)//n个人全做过，置0temp = 0;dp[i][temp] = max(dp[i][temp], dp[i - 1][j] + s[k + 1][i]);}}}}double ans = 0.0;for (int i = 0; i < (1 << n); i++)ans = max(ans, dp[m][i]);printf("Case #%d: %.5lf\n", cas, ans);}return 0;}