POJ3767 I Wanna Go Home dijkstra

题目链接：http://vjudge.net/contest/view.action?cid=47262#problem/A

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

2
1
1 2 100
1 2
3
3
1 2 100
1 3 40
2 3 50
1 2 1
5
5
3 1 200
5 3 150
2 5 160
4 3 170
4 2 170
1 2 2 2 1
0

Sample Output

100
90
540

 理解要点：
  1、永远不会走从领导2到领导1的路
   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

#include <iostream>
#include <cstdio>
#include <cstring>
#define inf  0x3fffffff
using namespace std;
int map[605][605];
int leader[605];
int dijks(int n)
{
int i,j,k,min;
int dis[605];
int vis[605];
memset(vis,0,sizeof(vis));
for(i=1;i<=n;i++){
dis[i]=map[1][i];//一到n的距离；
}
for(i=1;i<=n;i++){
min=inf;
k=1;
for(j=1;j<=n;j++){
if(!vis[j]&&min>dis[j]){
min=dis[j];
k=j;
}
}
vis[k]=1;
if(k==2)
break;
for( j=1;j<=n;j++){
if(!vis[j]&&dis[k]+map[k][j]<dis[j])
dis[j]=dis[k]+map[k][j];
}
}
if(dis[2]==inf)
return -1;
else
return dis[2];
}
int main()
{
int n,i,j,a,b,t,m;
while(~scanf("%d",&n),n)
{
for(i=1;i<=n;++i)
for(j=1;j<i;++j)
map[i][j]=map[j][i]=inf;
scanf("%d",&m);
while(m--)
{
scanf("%d%d%d",&a,&b,&t);
map[a][b]=map[b][a]=t;
}
for(i=1;i<=n;++i)
scanf("%d",&leader[i]);
for(i=1;i<=n;++i)
for(j=1;j<=n;++j)
if(leader[i]!=leader[j])//将领导者为从2~1的路变成不能走的路；
{
if(leader[i]==1)
map[j][i]=inf;
else
map[i][j]=inf;
}
printf("%d\n",dijks(n));
}
return 0;
}