Remmarguts‘ Date
| Time Limit: 4000MS | Memory Limit: 65536K | |
| Total Submissions: 26355 | Accepted: 7170 |
Description
"Good man never makes girls wait or breaks an appointment!" said the mandarin duck father. Softly touching his little ducks‘ head, he told them a story.
"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
The 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
A 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 思路：利用A*算法求最短路。第一次搜到终点为最短路，第二次搜到终点为次短路...第k次搜到终点为k短路。A*算法的估价函数f(x)=g(x)+h(x)。g(x)：从出发点到当前节点的距离。h(x)：从当前节点到终点的距离。可构建反向边利用spfa求得。
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
const int MAXN=1005;
const int INF=0x3f3f3f3f;
struct Edge{
int to,w,next;
}es[100005],rves[100005];
int n,m;
int src,tml,kth;
int head[MAXN],tot;
int rhead[MAXN],rtot;
void addedge(int u,int v,int w)
{
es[tot].to=v;
es[tot].w=w;
es[tot].next=head[u];
head[u]=tot++;
rves[rtot].to=u;
rves[rtot].w=w;
rves[rtot].next=rhead[v];
rhead[v]=rtot++;
}
int d[MAXN],vis[MAXN];
void spfa(int s)//利用反向边确定估价函数h(x):i到tml的距离
{
for(int i=1;i<=n;i++)
{
d[i]=INF;
vis[i]=0;
}
queue<int> que;
que.push(s);
d[s]=0;
vis[s]=1;
while(!que.empty())
{
int u=que.front();que.pop();
vis[u]=0;
for(int i=rhead[u];i!=-1;i=rves[i].next)
{
Edge e=rves[i];
if(d[e.to]>d[u]+e.w)
{
d[e.to]=d[u]+e.w;
if(!vis[e.to])
{
vis[e.to]=1;
que.push(e.to);
}
}
}
}
}
struct Node{
int nod,g,h;
Node(){}
Node(int cnod,int cg,int ch):nod(cnod),g(cg),h(ch){}
bool operator<(const Node &node) const
{
return g+h > node.g+node.h;
}
};
int astar(int s)
{
priority_queue<Node> que;
que.push(Node(s,0,d[s]));//g(x):src到i的距离.h(x):i到tml的距离.
while(!que.empty())
{
Node now=que.top();que.pop();
int u=now.nod;
if(u==tml)
{
kth--;
if(kth==0) return now.g+now.h;//第k条最短路
}
for(int i=head[u];i!=-1;i=es[i].next)
{
Edge e=es[i];
int g=now.g+e.w;
int h=d[e.to];
que.push(Node(e.to,g,h));
}
}
return -1;
}
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
memset(head,-1,sizeof(head));
memset(rhead,-1,sizeof(rhead));
tot=0;
rtot=0;
for(int i=0;i<m;i++)
{
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
addedge(u,v,w);
}
scanf("%d%d%d",&src,&tml,&kth);
if(src==tml) kth++;//若src等于tml那么最短路为0,不算
spfa(tml);
printf("%d\n",astar(src));
}
return 0;
}