Contest
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 220 Accepted Submission(s): 88
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
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 .
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.
2 3
0.6 0.3 0.4
0.3 0.7 0.9
#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 ;
}