HDOJ 3790 最短路径问题

题意：RT，单源最短路问题，求路径最短，当路径长相同时使花费尽可能少。

链接：http://acm.hdu.edu.cn/showproblem.php?pid=3790

SPFA/dijkstra模板即可

//cpp
//2014.09.19 18:05

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
using namespace std;

struct node{
    int en;
    int len;
    int val;
}adj[1005][1005];

int path[1005];
int val[1005];
int len[1005];

int SPFA ( int be, int en, int n )
{
    queue<int> p;
    p.push ( be );
    int tmp;
    while ( ! p.empty() )
    {
        tmp = p.front(); p.pop();
        for ( int i = 0; i < len[tmp]; i ++ )
        {
            if ( path[adj[tmp][i].en] > path[tmp] + adj[tmp][i].len )
            {
                path[adj[tmp][i].en] = path[tmp] + adj[tmp][i].len;
                val[adj[tmp][i].en]  = val[tmp]  + adj[tmp][i].val;
                p.push( adj[tmp][i].en );
            }
            else if ( path[adj[tmp][i].en] == path[tmp] + adj[tmp][i].len )
            {
                if ( val[adj[tmp][i].en] > val[tmp] + adj[tmp][i].val )
                {
                    path[adj[tmp][i].en] = path[tmp] + adj[tmp][i].len;
                    val[adj[tmp][i].en]  = val[tmp]  + adj[tmp][i].val;
                    p.push( adj[tmp][i].en );
                }
            }
        }
    }
}

int main()
{
    ios::sync_with_stdio( false );
    int m, n;
    while ( cin >> n >> m )
    {
        if ( ! m && ! n )
        {
            break;
        }
        int a, b, p, d;
        memset ( len, 0, sizeof ( len ) );
        memset ( adj, 0, sizeof ( adj ) );
        memset ( path, 0x3f3f3f3f, sizeof ( path ) );
        memset ( val,  0x3f3f3f3f, sizeof (  val ) );
        for ( int i = 0; i < m; i ++ )
        {
            cin >> a >> b >> d >> p;
            adj[a][len[a]].en = b;
            adj[a][len[a]].len = d;
            adj[a][len[a]].val = p;
            len[a] ++;
            adj[b][len[b]].en = a;
            adj[b][len[b]].len = d;
            adj[b][len[b]].val = p;
            len[b] ++;
        }
        int be, en;
        cin >> be >> en;
        path[be] = val[be] = 0;
        SPFA ( be, en, n );
        cout << path[en] << ' ' << val[en] << endl;
    }
    return 0;
}