2014上海网络赛1004||hdu5045 二分图的最佳匹配 或 状态压缩dp

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

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

   
   
   
   

    
    
    
    1
2 3
0.6 0.3 0.4
0.3 0.7 0.9
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case #1: 2.20000
   
   
   
   
   
   
   
   


   
   
   
   

这题数据只到10，学长是用状态压缩dp做的，自认为二分图写起来套费用流的模板时间更快一些。思路就是跑m/n+1次费用流，取和

#include<cstdio>
#include<iostream>
#include <string.h>
using namespace std;
const int oo=1e9;
const int mm=11111;
const int mn=888;
int node,src,dest,edge;
int ver[mm],flow[mm],next[mm];
double cost[mm],dis[mn];
int head[mn],p[mn],q[mn],vis[mn];

void prepare(int _node,int _src,int _dest)
{
    node=_node,src=_src,dest=_dest;
    for(int i=0; i<node; ++i)head[i]=-1,vis[i]=0;
    edge=0;
}
void addedge(int u,int v,int f,double c)
{
    ver[edge]=v,flow[edge]=f,cost[edge]=c,next[edge]=head[u],head[u]=edge++;
    ver[edge]=u,flow[edge]=0,cost[edge]=-c,next[edge]=head[v],head[v]=edge++;
}

bool spfa()
{
    int i,u,v,l,r=0;
    double tmp;
    for(i=0; i<node; ++i)
        dis[i]=oo;
    dis[q[r++]=src]=0;
    p[src]=p[dest]=-1;
    for(l=0; l!=r; (++l>=mn)?l=0:l)
        for(i=head[u=q[l]],vis[u]=0; i>=0; i=next[i])
            if(flow[i]&&dis[v=ver[i]]>(tmp=dis[u]+cost[i]))
            {
                dis[v]=tmp;
                p[v]=i^1;
                if(vis[v])continue;
                vis[q[r++]=v]=1;
                if(r>=mn)r=0;
            }
    return p[dest]>-1;
}

double SpfaFlow()
{
    int i,delta;
    double ret=0;
    while(spfa())
    {
        for(i=p[dest],delta=oo; i>=0; i=p[ver[i]])
            if(flow[i^1]<delta)delta=flow[i^1];
        for(i=p[dest]; i>=0; i=p[ver[i]])
            flow[i]+=delta,flow[i^1]-=delta;
        ret+=delta*dis[dest];
    }
    return ret;
}

int n,m,T;
double num[15][1005];

int main()
{
    int tt=0;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        memset(num,0,sizeof(num));
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
                scanf("%lf",&num[i][j]);
        double sum=0;
        for(int k=0;k<=m/n;k++)
        {
            prepare(n*2+2,0,n*2+1);
            for(int i=1;i<=n;i++)
            {
                addedge(src,i,1,0);
                addedge(i+n,dest,1,0);
                for(int j=1;j<=n;j++)
                    addedge(i,j+n,1,-num[i][j+k*n]);
            }
            sum+=(-SpfaFlow());
        }
        printf("Case #%d: %.5lf\n",++tt,sum);
    }
    return 0;
}

DP+状态压缩。dp[i][j]表示前i道题目j个人答题状态的最大值，j用二进制表示，因为人最多就10个。因为每两个人之间答题数目不能超过1，所以当状态达到1 << n - 1，即所有人都答过一题时，将重置为0。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;

int m,n;
double a[15][1250],dp[1205][1250];

int main()
{
    int T,tt=0;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;i++)
            for(int j=0;j<m;j++)
              scanf("%lf",&a[i][j]);
        for(int i=0;i<=m;i++)
            for(int j=0;j<1<<n;j++)
                dp[i][j]=-1.0;
        dp[0][0]=0;
        for(int i=0;i<m;i++)
        {
            for(int j=0;j<(1<<n);j++)
            {
                if(dp[i][j]<0)
                      continue;
                for(int k=0;k<n;k++)
                {
                    if(!((1<<k)&j))
                    {
                        int st=j|(1<<k);
                        if(st==(1<<n)-1)
                            st=0;
                        dp[i+1][st]=max(dp[i+1][st],dp[i][j]+a[k][i]);
                    }
                }
            }
        }
        /*for(int i=0;i<10;i++)
        {
            for(int j=0;j<10;j++)
                printf("%lf ",dp[i][j]);
            printf("\n");
        }*/
        double ans=0;
        for(int i=0;i<(1<<n);i++)
            ans=max(ans,dp[m][i]);
        printf("Case #%d: %.5lf\n",++tt,ans);
    }
    return 0;
}
/**

99
3 4
0.4 0.8 0.1 0.1
0.3 0.7 0.1 0.1
0.5 0.6 0.1 0.1

3 4
0.3 0.6 0.1 0.1
0.5 0.8 0.1 0.1
0.4 0.7 0.1 0.1

2 3
0.6 0.3 0.4
0.3 0.7 0.9

ANSWER:1.5,1.4,2.2
**/