九度OJ-1162：I Wanna Go Home

　　使用的依然是Dijkstra算法的模板。略作修改即可

题目描述：

    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?

输入：

    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.

输出：

    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.

样例输入：

211 2 1001 2331 2 1001 3 402 3 501 2 1553 1 2005 3 1502 5 1604 3 1704 2 1701 2 2 2 10

样例输出：

10090540

来源：

2011年北京大学计算机研究生机试真题

答疑：

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#include <iostream>#include <vector>#define MAXSIZE 600 #define MAXSIZE_EDGE 10000using namespace std;struct Edge{int start,end;double weight;Edge(){}Edge(int start,int end,double weight){this->start=start;this->end=end;this->weight=weight;}bool operator<(const Edge e){//uselessreturn weight<e.weight;}}; struct Vex{int root;Vex(){//initiate rootthis->root=-1;} };struct Graph{ //顶点下标从0开始 int graphSize;vector<Edge> edge[MAXSIZE_EDGE];int initGraph(int graphSize){this->graphSize=graphSize;for (int i=0;i<graphSize;i++)edge[i].clear();}};int main(){int n,m;Graph graph;int start,end,weight;bool mark[MAXSIZE];int dist[MAXSIZE];int conversion[MAXSIZE]; int support[MAXSIZE];int newPoint,minDistVex;int temp,tempCon;while (cin>>n,n){//n vexes ,m edgescin>>m;//initiategraph.initGraph(n);newPoint=0;//originate 0for (int i=0;i<graph.graphSize;i++){mark[i]=false;dist[i]=-1;conversion[i]=0;}mark[newPoint]=true;dist[newPoint]=0;//input edge&&supportfor (int i=0;i<m;i++){cin>>start>>end>>weight;start--;end--;graph.edge[start].push_back(Edge(start,end,weight)); graph.edge[end].push_back(Edge(end,start,weight));}for (int i=0;i<graph.graphSize;i++){cin>>support[i];}//processfor (int time=0;time<n-1;time++){//每趟循环找出x到一个结点的最短路径,共n-1趟 //遍历newPoint直接相邻的结点，修改其distfor (int i=0;i<graph.edge[newPoint].size();i++){int end=graph.edge[newPoint][i].end;if (mark[end]==true)//if end结点已经在x集合中，则跳过此次循环 continue;tempCon=conversion[newPoint]+(support[end]==support[newPoint]?0:1);//if 转向将要大于一次 continue if (tempCon>1)continue;temp=dist[newPoint]+graph.edge[newPoint][i].weight;if (temp<dist[end]||dist[end]==-1) {//if the new dist< the old dist,modify the pathdist[end]=temp;conversion[end]=tempCon;}}//遍历dist，从mark为false的结点中找出其dist最小的结点，确定为新的newPoint，并加入x集合 int i;for (i=0;i<graph.graphSize;i++){//initiate minDistVexif (mark[i]==false&&dist[i]!=-1){minDistVex=i;break;}}for (i++;i<graph.graphSize;i++){if (mark[i]==false&&dist[i]!=-1&&dist[i]<dist[minDistVex])minDistVex=i;} newPoint=minDistVex;mark[minDistVex]=true;} //output. output dist[]cout<<dist[1]<<endl;}return true;}