poj2387(Dijkstra)

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/qq_26760433/article/details/84486618

这题直接用Dijkstra算法就行了，没什么难度。要注意T是路径数，N是顶点数，别弄反了。可能会有重边的情况不过，不过我用的邻接表，没什么影响。
代码

//Menory 412k  Time 110ms
#include<vector>
#include<queue>
#include<iostream>
using namespace std;
int T,N;
class road
{
public:
	int end;
	int weight;
};
class Graph
{
public:
	int v;
	vector<road> *adj;
	int *mark;

	Graph(int n);
	~Graph();
	void addEdge(int s,int e,int w);
};
Graph::Graph(int n)
{
	v=n;
	adj=new vector<road>[n];
	mark=new int[n];
	for(int i=0;i<n;i++)
		mark[i]=0;
}
Graph::~Graph()
{
	delete []mark;
	delete []adj;
}
void Graph::addEdge(int s,int e,int w)
{
	road r;
	r.end=e;
	r.weight=w;
	adj[s].push_back(r);
	r.end=s;
	r.weight=w;
	adj[e].push_back(r);
}
class Dist
{
public:
	int index;
	int length;
	int pre;
	friend bool operator<(const Dist & a,const Dist &b)
	{
		return a.length>b.length;
	}
};
void Dijkstra(Graph & g,int s)
{
	Dist *D = new Dist[g.v];
	for(int i=0;i<g.v;i++)
	{
		D[i].index = i;
		D[i].length=2147483647;
		D[i].pre = s;
	}
	D[s].length = 0;
	priority_queue<Dist> aqueue;
	aqueue.push(D[s]);

	for(int i=0;i<g.v;i++)
	{
		Dist d;
		bool FOUND;
		FOUND=false;
		while(!aqueue.empty())
		{
			d=aqueue.top();
			aqueue.pop();
			if(g.mark[d.index]==0)
			{
				FOUND=true;
				break;
			}
		}
		if(!FOUND)
			break;
		g.mark[d.index]=1;
		int node=d.index;
		vector<road>::iterator ii=g.adj[node].begin();
		for(;ii!=g.adj[node].end();ii++)
		{
			if( D[ii->end].length> D[node].length + ii->weight)
			{
				D[ii->end].length = D[node].length + ii->weight;
				D[ii->end].pre=node;
				aqueue.push(D[ii->end]);
			}
		}
	}
	
	cout<<D[0].length<<endl;               //输出结果
}
int main()
{
	int a1,a2,a3;
	cin>>T>>N;
	Graph g(5000);
	for(int i=1;i<=T;i++)
	{
		cin>>a1>>a2>>a3;
		g.addEdge(a1-1,a2-1,a3);      //题目是从1开始，顶点减一从0开始
	}
	Dijkstra(g,N-1);
}