I want to go home

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?

输入描述:

    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.

输出描述:

    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.

示例1

输入

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

输出

100
90
540

//最短路径算法，通常情况下考察最短路径都会考察改进版。
//此题某人只能从1所领导的城市到2所了领导的城市，我们可以设置标记
#include<iostream>
#include<vector>
using namespace std;
struct E
{
    int next;//一个存储相邻的结点
    int cost;//存储边的权值
};


vector<E>edges[10001];
int dis[700];
int leader[700];
bool mark[701];//标记某顶点是否在最短路径顶点集合当中
int newp;
int main()
{
    int n,m;
    while(cin>>n>>m)
    {
        if(n==0)break;
        for(int i=1;i<=n;i++)
        {
            edges[i].clear();//每一个结点保存着其相邻的边
            mark[i]=false;
            dis[i]=-1;
        }
        for(int i=1;i<=m;i++)
        {
           E n;
           int a,b,cost;
           cin>>a>>b>>cost;
           n.next=b;
           n.cost=cost;
           edges[a].push_back(n);
           n.next=a;
           edges[b].push_back(n);
        }//这段是建立无向图的，任意一条边连接着两个顶点
        for(int i=1;i<=n;i++)
            cin>>leader[i];
        mark[1]=true;
        newp=1;
        dis[1]=0;//存储的权值（到第一个点权值和）
        for(int i=1;i<n;i++)//除了最后一个点
        {
            for(int j=0;j<edges[newp].size();j++)
            {
                int t=edges[newp][j].next;
                int c=edges[newp][j].cost;
                if(mark[t]==true)continue;//已经在路径集合当中
                if(leader[t]==1&&leader[newp]==2)continue;
                if(dis[t]==-1||dis[t]>c+dis[newp])
                {
                    dis[t]=dis[newp]+c;
                }
            }//找到newp到其每个相邻点的权值
            //求权值最小值并将那个点置为newp
                int minn=10000;
                for(int j=1;j<=n;j++)
                {
                    if(mark[j])continue;
                    if(dis[j]==-1)continue;
                    if(minn>dis[j])
                    {
                        minn=dis[j];
                        newp=j;
                    }
                }
                mark[newp]=true;
        }
        cout<<dis[2]<<endl;


    }
    return 0;


}