转载请注明出处： http://www.cnblogs.com/fraud/ ——by fraud
| Time Limit: 4000MS | Memory Limit: 65536K | |
| Total Submissions: 21855 | Accepted: 5958 |
Description
"Prince Remmarguts lives in his kingdom UDF – United Delta of
Freedom. One day their neighboring country sent them Princess Uyuw on a
diplomatic mission."
"Erenow, the princess sent Remmarguts a letter, informing him that
she would come to the hall and hold commercial talks with UDF if and
only if the prince go and meet her via the K-th shortest path. (in fact,
Uyuw does not want to come at all)"
Being interested in the trade development and such a lovely girl,
Prince Remmarguts really became enamored. He needs you - the prime
minister's help!
DETAILS: UDF's capital consists of N stations. The hall is numbered
S, while the station numbered T denotes prince' current place. M muddy
directed sideways connect some of the stations. Remmarguts' path to
welcome the princess might include the same station twice or more than
twice, even it is the station with number S or T. Different paths with
same length will be considered disparate.
Input
first line contains two integer numbers N and M (1 <= N <= 1000, 0
<= M <= 100000). Stations are numbered from 1 to N. Each of the
following M lines contains three integer numbers A, B and T (1 <= A, B
<= N, 1 <= T <= 100). It shows that there is a directed
sideway from A-th station to B-th station with time T.
The last line consists of three integer numbers S, T and K (1 <= S, T <= N, 1 <= K <= 1000).
Output
single line consisting of a single integer number: the length (time
required) to welcome Princess Uyuw using the K-th shortest path. If K-th
shortest path does not exist, you should output "-1" (without quotes)
instead.
Sample Input
2 2
1 2 5
2 1 4
1 2 2
Sample Output
14
题意：求第K短路
分析：spfa+A*
先spfa反向求最短路，然后根据A*来搞,f(x)=g(x)+h(x)
h(x)表示从终点反向到x点的最短距离，g(x)表示从起点到x的当前距离，在终点出队K次的时候所求的距离即为第K短路。
即我们每次都优先查找当前总的路程最短的路径，则在终点出队K次之后，即为第k短路了
#include <iostream>
#include <sstream>
#include <ios>
#include <iomanip>
#include <functional>
#include <algorithm>
#include <vector>
#include <string>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <set>
#include <map>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <climits>
#include <cctype>
using namespace std;
#define XINF INT_MAX
#define INF 0x3FFFFFFF
#define MP(X,Y) make_pair(X,Y)
#define PB(X) push_back(X)
#define REP(X,N) for(int X=0;X<N;X++)
#define REP2(X,L,R) for(int X=L;X<=R;X++)
#define DEP(X,R,L) for(int X=R;X>=L;X--)
#define CLR(A,X) memset(A,X,sizeof(A))
#define IT iterator
typedef long long ll;
typedef pair<int,int> PII;
typedef vector<PII> VII;
typedef vector<int> VI;
int s ,t,k;
const int maxn=;
vector<PII>G[maxn];
vector<PII>rG[maxn];
int dis[maxn];
int used[maxn];
void init(int n)
{
memset(used,,sizeof(used));
for(int i=;i<n;i++)
{
dis[i]=INF;
G[i].clear();
rG[i].clear();
}
}
void add_edge(int u,int v,int w){
G[u].push_back(make_pair(v,w));
rG[v].push_back(make_pair(u,w));
}
void spfa()
{
queue<int>q;
q.push(t);
used[t]=;
dis[t]=;
while(!q.empty())
{
int u=q.front();
for(int i=;i<rG[u].size();i++)
{
int v=rG[u][i].first;
int y=rG[u][i].second;
if(dis[u]+y<dis[v])
{
dis[v]=dis[u]+y;
if(!used[v])
{
used[v]=;
q.push(v);
}
}
}
q.pop();
used[u]=;
}
}
int A_star()
{
priority_queue<pair<int,PII>,vector<pair<int,PII> >,greater<pair<int,PII> > >q;
q.push(make_pair(dis[s],make_pair(,s)));
CLR(used,);
while(!q.empty())
{
pair<int,PII> p=q.top();
q.pop();
int f=p.first;
int g=p.second.first;
int u=p.second.second;
used[u]++;
if(used[t]==k)return f;
if(used[u]>k)continue;
for(int i=;i<G[u].size();i++)
{
int v=G[u][i].first;
int d=G[u][i].second;
q.push(make_pair(g+dis[v]+d,make_pair(g+d,v)));
}
}
return -;
}
int main()
{
ios::sync_with_stdio(false);
int n,m;
while(cin>>n>>m)
{
int u,v,w;
init(n);
for(int i=;i<m;i++)
{
cin>>u>>v>>w;
add_edge(--u,--v,w);
}
cin>>s>>t>>k;
s--;t--;
spfa();
if(s==t)k++;
cout<<A_star()<<endl;
}
return ;
}
代码君