HDU-3790 最短路最小花费

判断路径相等时的情况

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

#define pf printf
#define sf scanf
#define mp make_pair
#define INF 0x3f3f3f3f

const int MAXN = 1000 + 5;

int n,m;

struct node
{
    int d,fee;
}g[MAXN][MAXN];

int dis[MAXN],f[MAXN],vis[MAXN];


void dijkstra(int v)
{
    int i,j,mind,minf,dir;
    for(i=1;i<=n;i++)
    {
        dis[i] = f[i] = INF;
        vis[i] = 0;
    }
    dis[v] = f[v] = 0;
    vis[v] = 1;
    for(i=1;i<=n;i++)
    {
        mind = minf = INF;
        for(j=1;j<=n;j++)
        {
            if(!vis[j] && (dis[j]<mind || (dis[j]==mind && f[j]<minf)))
            {
                mind = dis[j];
                minf = f[j];
                dir = j;
            }
        }
        vis[dir] = 1;
        for(j=1;j<=n;j++)
        {
            if(!vis[j] && g[dir][j].d<INF && (dis[j] > (dis[dir]+g[dir][j].d) || (dis[j] == (dis[dir]+g[dir][j].d) && f[j]>f[dir]+g[dir][j].fee)))
            {
                dis[j] = dis[dir]+g[dir][j].d;
                f[j]=f[dir]+g[dir][j].fee;
            }
        }
    }

}


int main()
{
    int i,j;
    int a,b,d,fee,st,ed;
    while(sf("%d %d",&n,&m)==2,m+n)
    {
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=n;j++)
            {
                g[i][j].d = g[i][j].fee = (i==j ? 0:INF);
            }
        }
        for(i=0;i<m;i++)
        {
            sf("%d%d%d%d",&a,&b,&d,&fee);
            if(d < g[a][b].d)
            {
                g[a][b].d = g[b][a].d = d;
                g[a][b].fee = g[b][a].fee = fee;
            }
            else if(d == g[a][b].d)
                g[a][b].fee = g[b][a].fee = min(g[a][b].fee,fee);
        }
        sf("%d%d",&st,&ed);
        dijkstra(st);
        pf("%d %d\n",dis[ed],f[ed]);
    }
    return 0;