hdu 3790 最小费用最短路

求最小费用最短路：即在路径长度尽量小的情况下使得花费尽量小。

其实就是在松弛多了一种情况就是：路径长度相等但费用小。

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

const int INF = 99999999;
const int N = 1001;
const int M = 200000;
int head[N];
int dist[N];
int cost[N];
bool visit[N];
int n, m, e;

struct Edge
{
    int v, next, w, c;
} edge[M];

void addEdge( int u, int v, int w, int c )
{
    edge[e].v = v;
    edge[e].w = w;
    edge[e].c = c;
    edge[e].next = head[u];
    head[u] = e++;
}

void dij( int s )
{
    memset( visit, false, sizeof(visit) );
    for ( int i = 0; i <= n; i++ )
    {
        dist[i] = INF;
        cost[i] = INF;
    }
    dist[s] = 0;
    cost[s] = 0;
    for ( int i = 1; i <= n; i++ )
    {
        int u = 0;
        for ( int j = 1; j <= n; j++ )
        {
            if ( !visit[j] && dist[j] < dist[u] )
            {
                u = j;
            }
        }
        visit[u] = true;
        for ( int j = head[u]; j != -1; j = edge[j].next )
        {
            int v = edge[j].v, w = edge[j].w, c = edge[j].c;
            if ( visit[v] ) continue;
            if ( dist[u] + w < dist[v] )
            {
                dist[v] = dist[u] + w;
                cost[v] = cost[u] + c;
            }
            else if ( dist[u] + w == dist[v] && cost[u] + c < cost[v] )
            {
                cost[v] = cost[u] + c;
            }
        }
    }
}

int main ()
{
    while ( scanf("%d%d", &n, &m) != EOF )
    {
        if ( n == 0 && m == 0 ) break;
        e = 0;
        memset( head, -1, sizeof(head) );
        while ( m-- )
        {
            int u, v, w, c;
            scanf("%d%d%d%d", &u, &v, &w, &c);
            addEdge( u, v, w, c );
            addEdge( v, u, w, c );
        }
        int s, t;
        scanf("%d%d", &s, &t);
        dij(s);
        printf("%d %d
", dist[t], cost[t]);
    }
    return 0;
}