TLE in codeforce's Dijkstra
TLE in codeforces's Dijkstra
我刚刚学习了Dijkstra的算法,解决了一些问题,我正在努力解决这个问题http://codeforces.com/problemset/problem/20/C问题,但我在一个大型测试用例中遇到了超过时间限制的问题,我想知道我的代码是否可以以任何方式进一步优化,或者是否有任何其他更快的Dijkstra实现,然后告诉我。
我的代码
#include<bits/stdc++.h>
using namespace std;
#define pii pair<int,int>
#define vp vector<pii>
int p[100010],d[100010];
void printer(int current)
{
if(p[current]==-2)
{
printf("%d ",current);
return;
}
printer(p[current]);
printf("%d ",current);
}
class Prioritize
{
public:
int operator()(const pii &p1,const pii &p2)
{
return p1.second<p2.second;
}
};
int main()
{
priority_queue<pii, vp, Prioritize> Q;
int nv;
scanf("%d",&nv);
vp g[nv+1];
int ne,u,v,w;
scanf("%d",&ne);
for(int i=0;i<ne;i++)
{
scanf("%d %d %d",&u,&v,&w);
g[u].push_back(pii(v,w));
g[v].push_back(pii(u,w));
}
int source=1;
int size;
for(int i=1;i<=nv;i++)
{
d[i]=INT_MAX;
p[i]=-1;
}
d[source]=0;
p[source]=-2;//marker for source.
Q.push(pii(source,d[source]));
while(!Q.empty())
{
u=Q.top().first;
Q.pop();
size=g[u].size();
for(int i=0;i<size;i++)
{
v=g[u][i].first;
w=g[u][i].second;
if(d[v]>d[u]+w)
{
d[v]=d[u]+w;
p[v]=u;
Q.push(pii(v,d[v]));
}
}
}
/*for(int i=1;i<=nv;i++)
{
printf("Node %d and min weight = %d and parent = %dn",i,d[i],p[i]);
}*/
if(p[nv]==-1)
{
printf("%dn",-1);
return 0;
}
printer(nv);
return 0;
}
您的算法具有复杂性O(nm)
。我建议添加以下
Q.push(pii(source,d[source]));
while(!Q.empty())
{
u=Q.top().first;
int curD = Q.top().second; //this
if( curD > d[u]) continue; //and this
Q.pop();
size=g[u].size();
for(int i=0;i<size;i++)
{
变更后的复杂性将为O(mlogn)
。如果你现在是俄罗斯人,你会读http://e-maxx.ru/algo/dijkstra_sparse
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