hdu----(5045)Contest(数位dp)

Contest

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 220    Accepted Submission(s): 88

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 ith student solve the jth problem, the probability of correct solve is Pij .

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 jth number in the ith line is Pij .

 
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
1
2 3
0.6 0.3 0.4
0.3 0.7 0.9
 
Sample Output
Case #1: 2.20000
 
Source
 
 
数位DP:
代码:
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int maxn=1050l
double dp[maxn][maxn];
double a[][maxn];
int main(){
int cas,n,m;
scanf("%d",&cas);
for(int w=;w<=cas;w++){
scanf("%d%d",&n,&m);
for(int i=;i<=;i++){
for(int j=;j<=m;j++){
dp[i][j]=-;
}
}
dp[][]=0.0;
for(int i=;i<=n;i++){
for(int j=;j<=m;j++){
scanf("%lf",&a[i][j]);
}
}
int tol=(<<n)-;
double ans=0.0;
for(int j=;j<=m;j++){
for(int i=;i<tol;i++){
for(int k=;k<=n;k++){
int tt=<<(k-);
if(dp[i][j-]<)continue; if(tt&i)continue;
int pos=i|tt;
if(pos==tol)pos=;
dp[pos][j]=max(dp[pos][j],dp[i][j-]+a[k][j]);
if(j==m)ans=max(ans,dp[pos][j]);
}
}
}
printf("Case #%d: %.5lf\n",w,ans);
}
return ;
}
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