hdu 3499 Flight dijkstra 变形

4 4
Harbin Beijing 500
Harbin Shanghai 1000
Beijing Chengdu 600
Shanghai Chengdu 400
Harbin Chengdu

4 0
Harbin Chengdu

800
-1


Hint
In the first sample, Shua Shua should use the card on the flight from
 Beijing to Chengdu, making the route Harbin->Beijing->Chengdu have the
 least total cost 800. In the second sample, there‘s no way for him to get to 
Chengdu from Harbin, so -1 is needed. 
 

思路：分别以起点和终点为起点建立最短路，枚举打折的两个城市，求最短

#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <iostream>
#include <map>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;
const __int64 INF=1e18;
const int maxn=100010;
struct Edge{
	int from,to,dist;
	Edge(int u,int v,int d):from(u),to(v),dist(d) {}
};

struct HeapNode{
	int d,u;
	bool operator < (const HeapNode& rhs) const{
		return d > rhs.d;
	}
};
struct Dijkstra{
	int n,m;
	vector<Edge> edges;
	vector<int> G[maxn];
	bool done[maxn];
	__int64 d[maxn];
	
	void init(int n)
	{
		this->n=n;
		for(int i=0;i<n;i++) G[i].clear();
		edges.clear();
	}
	
	void addEdges(int from,int to,int dist)
	{
		edges.push_back(Edge(from,to,dist));
		m=edges.size();
		G[from].push_back(m-1);
	}
	
	
	
	void dijkstra(int s)
	{
		priority_queue<HeapNode> Q;
		for(int i=0;i<n;i++) d[i]=INF;
		d[s]=0;
		memset(done,0,sizeof(done));
		HeapNode tep;
		tep.d=0,tep.u=s;
		Q.push(tep);
		while (!Q.empty())
		{
			HeapNode x=Q.top();Q.pop();
			int u=x.u;
			if(done[u]) continue;
			done[u]=true;
			for(int i=0;i<G[u].size();i++)
			{
				Edge& e=edges[G[u][i]];
				if(d[e.to]>d[u]+e.dist)
				{
					d[e.to]=d[u]+e.dist;
					tep.d=d[e.to],tep.u=e.to;
					Q.push(tep);
				}
			}
		}
	}
	
	
};


string date;

__int64 temp;
map<string,int> data;
Dijkstra dij1,dij2;
int main ()
{
	__int64 ans;
	int tp,a,b,cnt,n,m;
	//freopen("data.in","r",stdin);
    while (~scanf("%d%d",&n,&m))
    {
		data.clear();
		dij1.init(n);
		dij2.init(n);
		cnt=-1;
        for (int i=1;i<=m;i++)
        {
            
            cin>>date;
			if(data.find(date)==data.end())
            	data[date]=++cnt;
			a=data[date];
            cin>>date;
			if(data.find(date)==data.end())
            	data[date]=++cnt;
			b=data[date];
            scanf("%d",&tp);
			dij1.addEdges(a,b,tp);
			dij2.addEdges(b,a,tp);
        }
        cin>>date;
		if(data.find(date)==data.end())
			data[date]=++cnt;
        a=data[date];
		cin>>date;
		if(data.find(date)==data.end())
			data[date]=++cnt;
        b=data[date];
		dij1.dijkstra(a);
		dij2.dijkstra(b);
        ans=INF;
		for(int i=0;i<dij1.edges.size();i++)
		{
			Edge tem=dij1.edges[i];
			temp=dij1.d[tem.from]+dij2.d[tem.to]+tem.dist/2;
			ans=ans<temp?ans:temp;
		}
        if(ans>=INF) puts("-1");
        else printf("%I64d\n",ans);
    }
}

hdu 3499 Flight dijkstra 变形,布布扣,bubuko.com

hdu 3499 Flight dijkstra 变形

原文：http://blog.csdn.net/alpc_paul/article/details/38495959