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如果会Dijstra算法,这题水水的。就是题意有点难懂。
Sample Input
3 2 1
2 2 3
2 3 1
2 1 2
第一行,表示有3个顶点,求从2走向1的最短路径。
第二行,表示从顶点1发出两条边,默认通向顶点2路径长度为0,通向顶点3路径长度为1.(不是默认通向的顶点路径长度为1)
第三行,表示从顶点2发出两条边,默认通向顶点3路径长度为0,通向顶点1路径长度为1.
第四行,表示从顶点3发出两条边,默认通向顶点1路径长度为0,通向顶点2路径长度为1.
最忧路径为 2->3->1
代码
#include<vector>
#include<queue>
#include<iostream>
using namespace std;
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);
}
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,int e)
{
Dist *D = new Dist[g.v];
for(int i=0;i<g.v;i++)
{
D[i].index = i;
D[i].length=1<<30;
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]);
}
}
}
if(D[e].length==1<<30)
cout<<"-1"<<endl;
else
cout<<D[e].length<<endl;
}
int main()
{
int begin,end;
int num_node;
int temp;
int flag;
int now_end;
cin>>num_node>>begin>>end;
begin=begin-1;
end=end-1;
Graph g(105);
for(int i=0;i<num_node;i++)
{
flag=0;
cin>>temp;
while(temp--)
{
cin>>now_end;
if(flag==0)
{
g.addEdge(i,now_end-1,0);
flag=1;
}
else
g.addEdge(i,now_end-1,1);
}
}
Dijkstra(g,begin,end);
}