POJ 2449 Remmarguts' Date ( Dijkstra + A* 求解第K短路 )

Remmarguts' Date

Time Limit: 4000MS Memory Limit: 65536KTotal Submissions: 21855 Accepted: 5958

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

Source

POJ Monthly,Zeyuan Zhu

#include <iostream>#include <cstring>#include <queue>#include <fstream>using namespace std;#define E 100005#define V 1005#define INF 1 << 30int heads[V], r_heads[V];int dists[V];bool visits[V];int nEdgeNum, nNodeNum, nEdgeCount;int nEnd, nSrc, k;struct Edge{    int to_node;    int next_edge;    int edge_weight;    int r_to_node;    int r_next_edge;    Edge(){}    Edge( int from, int to, int weight ){        to_node     = to;        r_to_node   = from;        edge_weight = weight;     }}edges[E];struct Node{    int v;    int src_to_v_dist;    Node(){        this->v             = 0;        this->src_to_v_dist = 0;    }    Node( int v, int d ){        this->v             = v;        this->src_to_v_dist = d;    }    bool operator < ( const Node& other ) const{        return src_to_v_dist + dists[v] > dists[other.v] + other.src_to_v_dist;    }};void addEdge( int from, int to, int dist ){    edges[nEdgeCount]             = Edge( from, to, dist );    edges[nEdgeCount].r_next_edge = r_heads[to];    edges[nEdgeCount].next_edge   = heads[from];    heads[from]                   = nEdgeCount;    r_heads[to]                   = nEdgeCount;    nEdgeCount++;}void dijkstra( int src ){    priority_queue< Node > que;    for( int i = 1; i <= nNodeNum; ++i )        dists[i] = INF;    dists[src] = 0;    que.push( Node( src, 0 ) );    while( !que.empty() ){        Node cur = que.top();        que.pop();        if( visits[cur.v] )            continue;        visits[cur.v] = true;        for( int i = r_heads[cur.v]; ~i; i = edges[i].r_next_edge ){            if( dists[edges[i].r_to_node] > dists[cur.v] + edges[i].edge_weight ){                dists[edges[i].r_to_node] = dists[cur.v] + edges[i].edge_weight;                que.push( Node( edges[i].r_to_node, 0 ) );            }        }    }}int AStar( int src ){    priority_queue< Node > que;    que.push( Node( src, 0 ) );    while( !que.empty() ){        Node cur = que.top();        que.pop();        if( cur.v == nEnd ){            if( k > 1 )                k--;            else                return cur.src_to_v_dist;        }        for( int i = heads[cur.v]; ~i; i = edges[i].next_edge )            que.push( Node( edges[i].to_node, cur.src_to_v_dist + edges[i].edge_weight ) );    }    return -1;}void init(){    memset( visits,  false, sizeof( visits ) );    memset( heads,   -1,    sizeof( heads ) );    memset( r_heads, -1,    sizeof( r_heads ) );    nEdgeCount = 0;}int main(){    //fstream fin( "t.txt" );    while( cin >> nNodeNum >> nEdgeNum ){        init();        for( int i = 1; i <= nEdgeNum; ++i ){            int from, to, dist;            cin >> from >> to >> dist;            addEdge( from, to, dist );        }        cin >> nSrc >> nEnd >> k;        dijkstra( nEnd );        if( dists[nSrc] == INF ){            cout << "-1\n";            continue;        }        if( nSrc == nEnd )            k++;        int ans = AStar( nSrc );        cout << ans << endl;    }    return 0;}