Remmarguts‘ Date
| Time Limit: 4000MS | Memory Limit: 65536K | |
| Total Submissions: 22412 | Accepted: 6085 |
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
Source
POJ Monthly,Zeyuan Zhu
题目链接：http://poj.org/problem?id=2449
题目大意：n个点，m条路，给出起点s和终点t，求从s到t的第k短路的长度
题目分析：A*算法模板题
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
using namespace std;
int const INF = 0xfffffff;
int const MAX1 = 1005;
int const MAX2 = 1e5 + 5;
int dis[MAX1], head[MAX1], head2[MAX1];
int n, m, cnt;
bool vis[MAX1];
struct node
{
int v, w;
int next;
}e[MAX2], e2[MAX2];
struct node1
{
int v, g, f;
bool operator < (const node1 &r) const
{
if(r.f == f)
return r.g < g;
return r.f < f;
}
};
void Add(int u, int v, int w)
{
e[cnt].v = v;
e[cnt].w = w;
e[cnt].next = head[u];
head[u] = cnt;
e2[cnt].v = u;
e2[cnt].w = w;
e2[cnt].next = head2[v];
head2[v] = cnt++;
}
bool spfa(int s)
{
memset(vis, false, sizeof(vis));
dis[s] = 0;
queue <int> q;
q.push(s);
while(!q.empty())
{
int cur = q.front();
q.pop();
vis[cur] = false;
for(int i = head2[cur]; i != -1; i = e2[i].next)
{
if(dis[e2[i].v] > dis[cur] + e2[i].w)
{
dis[e2[i].v] = dis[cur] + e2[i].w;
if(!vis[e2[i].v])
{
vis[e2[i].v] = true;
q.push(e2[i].v);
}
}
}
}
}
int A_star(int s, int t, int k)
{
if(s == t)
k++;
if(dis[s] == INF)
return -1;
priority_queue <node1> q;
int num = 0;
node1 tmp, to;
tmp.v = s;
tmp.g = 0;
tmp.f = tmp.g + dis[tmp.v];
q.push(tmp);
while(!q.empty())
{
tmp = q.top();
q.pop();
if(tmp.v == t)
num++;
if(num == k)
return tmp.g;
for(int i = head[tmp.v]; i != -1; i = e[i].next)
{
to.v = e[i].v;
to.g = tmp.g + e[i].w;
to.f = to.g + dis[to.v];
q.push(to);
}
}
return -1;
}
int main()
{
scanf("%d %d", &n, &m);
cnt = 0;
memset(head, -1, sizeof(head));
memset(head2, -1, sizeof(head2));
for(int i = 0; i < MAX1; i++)
dis[i] = INF;
while(m--)
{
int u, v, w;
scanf("%d %d %d", &u, &v, &w);
Add(u, v, w);
}
int s, t, k;
scanf("%d %d %d", &s, &t, &k);
spfa(t);
printf("%d\n", A_star(s, t, k));
}
POJ 2449 Remmarguts' Date (第k短路 A*搜索算法模板)