UVA 10917(p330)----Walk through the Forest

#include<bits/stdc++.h>
#define debu
using namespace std;
const int INF=1e10;
const int maxn=1e3+50;
struct Edge
{
int from,to,dist;
Edge(int a=0,int b=0,int c=0):from(a),to(b),dist(c) {}
};
struct HeapNode
{
int d,u;
HeapNode(int a=0,int b=0):d(a),u(b) {}
bool operator < (const HeapNode& rhs) const
{
return d>rhs.d;
}
};
int d[maxn];
vector<Edge> edges;
vector<int> G[maxn];
struct Dijkstra
{
int n,m;
bool done[maxn];
int p[maxn];
void init(int n)
{
this->n=n;
for(int i=0; i<n; i++) G[i].clear();
edges.clear();
}
void AddEdge(int from,int to,int dist)
{
edges.push_back(Edge(from,to,dist));
int 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));
q.push(HeapNode(0,s));
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;
p[e.to]=G[u][i];
q.push(HeapNode(d[e.to],e.to));
}
}
}
}
};
int n,m;
int f[maxn];
Dijkstra ans;
vector<int> g[maxn];
void make()
{
// cout<<edges.size()<<endl;
/* for(int i=0;i<n;i++)
cout<<i<<" "<<d[i]<<endl;*/
for(int i=0;i<edges.size();i++)
{
Edge e=edges[i];
// cout<<e.from<<" "<<e.to<<endl;
int u=e.from,v=e.to;
if(d[v]<d[u])
{
//  cout<<"flag "<<u<<" "<<v<<endl;
g[u].push_back(v);
}
}
}
int solve(int u)
{
int sum=0;
if(f[u]!=-1) return f[u];
for(int i=0; i<g[u].size(); i++)
sum+=solve(g[u][i]);
return f[u]=sum;
}
int main()
{
#ifdef debug
freopen("in.in","r",stdin);
#endif // debug
while(scanf("%d%d",&n,&m)==2&&n)
{
ans.init(n);
for(int i=0; i<n; i++) g[i].clear();
memset(f,-1,sizeof(f));
for(int i=0; i<m; i++)
{
int x,y,w;
scanf("%d%d%d",&x,&y,&w);
x--;y--;
ans.AddEdge(x,y,w);
ans.AddEdge(y,x,w);
}
ans.dijkstra(1);
make();
f[1]=1;
solve(0);
printf("%d\n",f[0]);
}
return 0;
}

题目地址：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1858

题解：对于题目规定路径，即指d[B]<d[A]，因此建立一个新图，当且仅当d[B]<d[A]时加入有向边A->B，则ans等于新图中起点到终点的路径条数。f[u]=sum(f[v])（v为u的后继结点,f[1]=1)记忆化搜索即可。