POJ3311 Hie with the Pie

Posted 2020-09-26 嘒彼小星

Hie with the Pie

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 7624		Accepted: 4106

Description

The Pizazz Pizzeria prides itself in delivering pizzas to its customers as fast as possible. Unfortunately, due to cutbacks, they can afford to hire only one driver to do the deliveries. He will wait for 1 or more (up to 10) orders to be processed before he starts any deliveries. Needless to say, he would like to take the shortest route in delivering these goodies and returning to the pizzeria, even if it means passing the same location(s) or the pizzeria more than once on the way. He has commissioned you to write a program to help him.

Input

Input will consist of multiple test cases. The first line will contain a single integer n indicating the number of orders to deliver, where 1 ≤ n ≤ 10. After this will be n + 1 lines each containing n + 1 integers indicating the times to travel between the pizzeria (numbered 0) and the n locations (numbers 1 to n). The jth value on the ith line indicates the time to go directly from location i to location j without visiting any other locations along the way. Note that there may be quicker ways to go from i to j via other locations, due to different speed limits, traffic lights, etc. Also, the time values may not be symmetric, i.e., the time to go directly from location i to j may not be the same as the time to go directly from location j to i. An input value of n = 0 will terminate input.

Output

For each test case, you should output a single number indicating the minimum time to deliver all of the pizzas and return to the pizzeria.

Sample Input

3
0 1 10 10
1 0 1 2
10 1 0 10
10 2 10 0
0

Sample Output

8

Source

East Central North America 2006

 

 

大意如下：

给定一张有n + 1个节点的有向图，每两个点之间均有两条方向不一的边，权值不一定相同。问你从0号节点出发，走过所有1...n个节点并回到0号节点所花费的时间。

 

首先Floyd预处理多源最短路。

 

状压方程：

f[S][i]表示从0号点出发，走过集合S中的点并到达i点的最短路长度。

 

转移：

f[S][i] = f[S/i][j] + g[j][i];

 

初始状态:

f[(1 << (i - 1))][i] = g[0][i]

 

答案:

min(f[(1 << n) - 1][i] + g[i][0]),i为1....n

 

Code

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <cstdlib>
 4 #include <cstring>
 5 #include <algorithm>
 6 #include <vector>
 7 #include <queue>
 8 #include <stack>
 9 inline void read(int &x){x = 0;char ch = getchar();char c = ch;while(ch < ‘0‘ || ch > ‘9‘)c = ch, ch = getchar();while(ch <= ‘9‘ && ch >= ‘0‘)x = x * 10 + ch - ‘0‘, ch = getchar();if(c == ‘-‘)x = -x;}
10 inline int max(int a, int b){return a > b ? a : b;}
11 inline int min(int a, int b){return a < b ? a : b;}
12 
13 const int INF = 0x3f3f3f3f;
14 const int MAXN = 10 + 4;
15 
16 int n,g[MAXN][MAXN],dp[1 << MAXN][MAXN],ans;
17 
18 inline void init()
19 {
20     //清空
21     memset(g, 0, sizeof(g));
22     memset(dp, 0, sizeof(dp));
23     //读入
24     for(int i = 0;i <= n;++ i)
25         for(int j = 0;j <= n;j ++)
26             read(g[i][j]);
27 }
28 
29 inline void Floyd()
30 {
31     for(int k = 0;k <= n;++ k)
32         for(int i = 0;i <= n;++ i)
33             for(int j = 0;j <= n;++ j)
34                 if(g[i][j] > g[i][k] + g[k][j])
35                     g[i][j] = g[i][k] + g[k][j];
36 }
37 
38 //dp[S][j]  表示已经经过点集i到达j点的最短时间   1表示经过，0表示未经过
39 //dp[S][j] = min(dp[S/k][k] + g[k][j])  k∈S  iS/k表示i去掉k 
40 
41 inline void DP()
42 {    
43     //枚举点集 
44     for (int S = 1;S < (1 << n); ++ S)
45         //枚举出发点 
46         for (int i = 1;i <= n;++ i)
47         {
48             if(!(S & (1 << (i - 1))))continue; 
49             //如果集合中只有出发点一个点  到达它需要g[0][i]的时间 
50             if (S == (1 << (i - 1)))
51             {
52                 dp[S][i] = g[0][i];
53                 break;
54             }
55             dp[S][i] = INF;
56             for (int j = 1;j <= n;++ j)
57             {
58                 if (j == i)continue;
59                 if (!(S & (1 << (j - 1))))continue;
60                 dp[S][i] = min(dp[S][i], dp[S ^ (1 << (i - 1))][j] + g[j][i]);
61             }    
62         }
63     ans = INF;
64     for (int i = 1;i <= n;++ i)
65         ans = min(ans, dp[(1 << n) - 1][i] + g[i][0]);
66 }
67 
68 inline void out()
69 {
70     printf("%d\n", ans);
71 }
72 
73 int main()
74 {
75     while (scanf("%d", &n))
76     {
77         if (!n)break;
78         //初始化操作 
79         init(); Floyd();DP(); out();
80     }
81     return 0;
82 }

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