POJ-3767IWannaGoHome-----有限制的dijkstra

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I Wanna Go Home

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2481		Accepted: 1035

Description

The country is facing a terrible civil war----cities in the country are divided into two parts supporting different leaders. As a merchant, Mr. M does not pay attention to politics but he actually knows the severe situation, and your task is to help him reach home as soon as possible.

"For the sake of safety,", said Mr.M, "your route should contain at most 1 road which connects two cities of different camp."

Would you please tell Mr. M at least how long will it take to reach his sweet home?

Input

The input contains multiple test cases.

The first line of each case is an integer N (2<=N<=600), representing the number of cities in the country.

The second line contains one integer M (0<=M<=10000), which is the number of roads.

The following M lines are the information of the roads. Each line contains three integers A, B and T, which means the road between city A and city B will cost time T. T is in the range of [1,500].

Next part contains N integers, which are either 1 or 2. The i-th integer shows the supporting leader of city i.

To simplify the problem, we assume that Mr. M starts from city 1 and his target is city 2. City 1 always supports leader 1 while city 2 is at the same side of leader 2.

Note that all roads are bidirectional and there is at most 1 road between two cities.

Input is ended with a case of N=0.

Output

For each test case, output one integer representing the minimum time to reach home.

If it is impossible to reach home according to Mr. M's demands, output -1 instead.

Sample Input

Sample Output

100
90
540

 1 /* 功能Function Description:     POJ-3767  ----有限制的dijkstra
 2    开发环境Environment:          DEV C++ 4.9.9.1
 3    技术特点Technique:
 4    版本Version:
 5    作者Author:                   可笑痴狂
 6    日期Date:                      20120821
 7    备注Notes:
 8         这道题想清楚了很简单，但是开始时候还是迷茫很久。
 9         理解要点：
10         1、永远不会走从领导2到领导1的路
11         2、先根据题意建立双向路，然后把存在领导2到领导1的路去除，让其变成领导1到领导2的单项路，之后就是dijkstra了
12 */
13 #include<stdio.h>
14 #include<string.h>
15 #define inf 0x3fffffff
16 int map[605][605];
17  
18 int dijks(int n)
19 {
20     int i,j,k,min;
21     int dis[605],visit[605];
22     memset(visit,0,sizeof(visit));
23     visit[1]=1;
24     for(i=1;i<=n;++i)
25         dis[i]=map[1][i];
26     for(i=1;i<n;++i)
27     {
28         min=inf;
29         k=1;
30         for(j=1;j<=n;++j)
31         {
32             if(!visit[j]&&min>dis[j])
33             {
34                 min=dis[j];
35                 k=j;
36             }
37         }
38         visit[k]=1;
39         if(k==2)
40             break;
41         for(j=1;j<=n;++j)
42         {
43             if(!visit[j]&&dis[k]+map[k][j]<dis[j])
44                 dis[j]=dis[k]+map[k][j];
45         }
46     }
47     if(dis[2]==inf)
48         return -1;
49     else
50         return dis[2];
51 }
52 
53 int main()
54 {
55     int n,i,j,a,b,t,m;
56     int leader[605];
57     while(scanf("%d",&n),n)
58     {
59         for(i=1;i<=n;++i)
60             for(j=1;j<i;++j)
61                 map[i][j]=map[j][i]=inf;
62         scanf("%d",&m);
63         while(m--)
64         {
65             scanf("%d%d%d",&a,&b,&t);
66             map[a][b]=map[b][a]=t;
67         }
68         for(i=1;i<=n;++i)
69             scanf("%d",&leader[i]);
70         for(i=1;i<=n;++i)
71             for(j=1;j<=n;++j)
72                 if(leader[i]!=leader[j])
73                 {
74                     if(leader[i]==1)
75                         map[j][i]=inf;
76                     else
77                         map[i][j]=inf;
78                 }
79         printf("%d\n",dijks(n));
80     }
81     return 0;
82 }