HOJ-10513 Allocation Scheme[简单DFS]

http://acm.hnu.cn/online/?action=problem&type=show&id=10513

Allocation Scheme
Time Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:32768KB
Total submit users: 69, Accepted users: 64
Problem 10513 : No special judgement
Problem description
A manager of one company received a emergent project that needs to be completed in shortest time。With considering and analysing, the manager would divide the project into N independent tasks and that needs N employees to complete. Every ernployee can do any one of the N tasks, but the time is different. Please design a allocation scheme for the manager to make the task can be completed in shortest time.

Input
The number of the employees N begins with 0, so is the tasks number N. The time of every task done by every employee is stored in a two-dimensional array task_worker[N][N]. For example: task_worker[i][j] means the time of task i completed by employee j.
Output
The first row show the shortest time to complete the project.(unit: hour)
Output the situation of the allocation scheme.

Sample Input 
10  11  12  11  9  11
11  9   10  13  11  12
12  10  11  10  13  9
9   14  9   10  10  11
10  10  9   11  12  11
10  7  10   10  10   8
Sample Output 
The shortest time is 54 hours
Task 0 is distributed to employee 4
Task 1 is distributed to employee 1
Task 2 is distributed to employee 3
Task 3 is distributed to employee 0
Task 4 is distributed to employee 2
Task 5 is distributed to employee 5
Problem Source

HNU Contest 

 

 

 

注意一下输入，要以文件结束。

code:

 1  #include <iostream>   
 2  #include <iomanip>   
 3  #include <fstream>   
 4  #include <sstream>   
 5  #include <algorithm>   
 6  #include <string>   
 7  #include <set>   
 8  #include <utility>   
 9  #include <queue>   
10  #include <stack>   
11  #include <list>   
12  #include <vector>   
13  #include <cstdio>   
14  #include <cstdlib>   
15  #include <cstring>   
16  #include <cmath>   
17  #include <ctime>   
18  #include <ctype.h> 
19  using namespace std;
20  
21  int map[101][101];
22  char str[101];
23  int record[101];
24  int record1[101];
25  int vst[101];
26  int mintime;
27  int n;
28  char t[101]; 
29  
30  void DFS(int cnt,int sum)
31  {
32      int i;
33      if(cnt==n+1)
34      {
35          if(sum<mintime)
36          {
37              mintime=sum;
38              for(i=0;i<=n;i++)
39                  record1[i]=record[i];
40          }
41          return;  
42     }
43     for(i=0;i<=n;i++)
44     {
45         if(!vst[i])
46         {
47             vst[i]=1;
48             record[cnt]=i; 
49             DFS(cnt+1,sum+map[cnt][i]);
50             vst[i]=0;
51         }
52     }
53  }
54  
55  int main()
56  {
57      int i,j;
58      int temp;
59      while(gets(str))
60      {
61          memset(map,0,sizeof(map));
62          memset(record,0,sizeof(record));
63          n=0;
64          temp=0;
65          for(i=0;i<strlen(str);i++)
66          {
67              if(str[i]==' '&&str[i+1]!=' ')
68              {
69                  map[0][n]/=10;
70                  n++;
71              }
72             else if(str[i]!=' ') 
73             {
74                 map[0][n]+=(int(str[i])-48);
75                 map[0][n]*=10;
76             } 
77             else{
78                 
79             }
80          }
81          for(i=1;i<=n;i++)
82              for(j=0;j<=n;j++)
83                  scanf("%d",&map[i][j]);
84          cin>>t; 
85        mintime=9999999;
86        for(i=0;i<=n;i++)
87        {
88            memset(vst,0,sizeof(vst));
89            memset(record,0,sizeof(record));
90              vst[i]=1;
91              record[0]=i;
92              DFS(1,map[0][i]);
93        }
94        printf("The shortest time is %d hours\n",mintime);
95        for(i=0;i<=n;i++)
96            printf("Task %d is distributed to employee %d\n",i,record1[i]);
97     }
98     return 0;
99  }

 

 

posted @ 2012-07-20 09:06  max_xbw  阅读(265)  评论(0)  编辑  收藏  举报