九度 OJ 1162 I Wanna Go Home 最短路問題

題目連結：http://ac.jobdu.com/problem.php?pid=1162

題目描述：

    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?

輸入：

    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.

輸出：

    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.

樣例輸入：

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      

樣例輸出：

100
90
540      

     此題屬于單源最短路徑問題，隻是建構圖時要做一些小改動，因為主人公不想來回穿越兩軍的地盤，有要從1到2，是以要想辦法把陣營2到陣營1的路徑設為不可到達，我在這是把邊權值設為0辦法，求最短路時判0跳過。圖建構完之後再用Dijsktra最短路算法求得最短路。

#include <stdio.h>
#include <vector>
using namespace std;
struct E{
	int next;
	int t;
};
vector <E> edge[610];
bool mark[610];
int time[610];
int owner[610];
int main(){
	int n,m;
	while(scanf("%d",&n)!=EOF && n!=0){
		scanf("%d",&m);
		for(int i=1;i<=n;i++){
			edge[i].clear();
		}
		int a,b,t;
		for(int i=0;i<m;i++){
			scanf("%d%d%d",&a,&b,&t);
			E tmp;
			tmp.next=b;
			tmp.t=t;
			edge[a].push_back(tmp);
			tmp.next=a;
			edge[b].push_back(tmp);
		}
		for(int i=1;i<=n;i++){
			mark[i]=false;
			time[i]=-1;
			scanf("%d",&owner[i]); 
		}
		for(int i=1;i<=n;i++){
			if(owner[i]==2){
				for(int j=0;j<edge[i].size();j++){
					if(owner[edge[i][j].next]==1){
						edge[i][j].t=0;
					}
				}
			}
		}
		time[1]=0;
		mark[1]=true;
		int newp=1;
		for(int i=1;i<n;i++){
			for(int j=0;j<edge[newp].size();j++){
				int next_=edge[newp][j].next;
				int time_=edge[newp][j].t;
				if(mark[next_]==true)
					continue;
				if(time_==0)
					continue;
				if(time[next_]==-1 || time[next_]>time[newp]+time_){
					time[next_]=time[newp]+time_;
				}
			}
				int min=1000;
				for(int j=1;j<=n;j++){
					if(mark[j]==true)
						continue;
					if(time[j]==-1)
						continue;
					if(time[j]<min){
						min=time[j];
						newp=j;
					}
				}
				mark[newp]=true;
			}
			if(time[2]!=-1)
				printf("%d\n",time[2]);
			else
				printf("-1\n");
	}
	return 0;
}