POJ2449 Remmarguts' Date 【k短路】

Remmarguts' Date

Time Limit: 4000MS Memory Limit: 65536KTotal Submissions: 21064 Accepted: 5736

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 21 2 52 1 41 2 2

Sample Output

14

第一次做k短路的题，大致总结下这个算法：

1、首先将有向图的每条边倒置构成一个新图，然后求终点到每个顶点的最短距离（SPFA或Dijkstra都行）；

2、定义结构体类型，变量包括链接的终点，已走路径长度g和估价函数值f，f在优先队列的队首开始升序排列；

3、定义cnt,表示当前已到达终点的次数，若cnt==k,算法完成，另外若起点跟终点重合的话应该++k。

#include <stdio.h>#include <string.h>#include <queue>#define maxn 1002#define maxm 100002#define inf 0x7fffffffusing std::priority_queue;using std::queue;int head[maxn], Rhead[maxn]; //Rhead[] save the reverse Graphstruct Node{int to, dist, next;} E[maxm], RE[maxm];int dist[maxn]; //reversestruct  Node2{int to, f, g; //估价函数和已走路径长度bool operator<(const Node2& a) const{return a.f < f;}};bool vis[maxn];void addEdge(int u, int v, int d, int i){E[i].to = v; E[i].dist = d;E[i].next = head[u]; head[u] = i;RE[i].to = u; RE[i].dist = d;RE[i].next = Rhead[v]; Rhead[v] = i;}void SPFA(int s, int n, int dist[], int head[], Node E[]){int i, u, v, tmp;for(i = 0; i <= n; ++i){dist[i] = inf; vis[i] = false;}u = s; vis[u] = true; dist[u] = 0;queue<int> Q; Q.push(s);while(!Q.empty()){u = Q.front(); Q.pop(); vis[u] = 0;for(i = head[u]; i != -1; i = E[i].next){v = E[i].to;tmp = dist[u] + E[i].dist;if(tmp < dist[v]){dist[v] = tmp;if(!vis[v]){vis[v] = 1;Q.push(v);}}}}}int A_star(int s, int t, int k, int n){priority_queue<Node2> Q;int i, u, v, cnt = 0;Node2 tmp, now;if(s == t) ++k; //起点跟终点相同now.to = s; now.g = 0;now.f = now.g + dist[now.to];Q.push(now);while(!Q.empty()){now = Q.top(); Q.pop();if(now.to == t && ++cnt == k)return now.g;for(i = head[now.to]; i != -1; i = E[i].next){tmp.to = E[i].to; tmp.g = now.g + E[i].dist;tmp.f = tmp.g + dist[tmp.to]; Q.push(tmp);}}return -1;}int main(){int n, m, s, t, k, u, v, d, i;while(scanf("%d%d", &n, &m) == 2){for(i = 0; i <= n; ++i)head[i] = Rhead[i] = -1;for(i = 0; i < m; ++i){scanf("%d%d%d", &u, &v, &d);addEdge(u, v, d, i);}scanf("%d%d%d", &s, &t, &k);//SPFA(s, n, dist, head, E);SPFA(t, n, dist, Rhead, RE);printf("%d\n", A_star(s, t, k, n));}return 0;}