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题意：给出n个点m条无向边，每条边有一个距离和一个花费。给出s,t。问从s到t的最短距离以及最短距离时的最小花费。当有多个距离最短的方案时，选取花费最少的。
先写几道题加深理解。
记得初始化。。。。。。
代码实现
1/* ***********************************************
2Author :111qqz
3Created Time :2016年05月21日 星期六 18时42分24秒
4File Name :code/hdu/3790.cpp
5************************************************ */
6
7#include <cstdio>
8#include <cstring>
9#include <iostream>
10#include <algorithm>
11#include <vector>
12#include <queue>
13#include <set>
14#include <map>
15#include <string>
16#include <cmath>
17#include <cstdlib>
18#include <ctime>
19#define fst first
20#define sec second
21#define lson l,m,rt<<1
22#define rson m+1,r,rt<<1|1
23#define ms(a,x) memset(a,x,sizeof(a))
24typedef long long LL;
25#define pi pair < int ,int >
26#define MP make_pair
27
28using namespace std;
29const double eps = 1E-8;
30const int dx4[4]={1,0,0,-1};
31const int dy4[4]={0,-1,1,0};
32const int inf = 0x3f3f3f3f;
33const int N=1E3+7;
34int n,m;
35int d[N];
36int p[N];
37bool inq[N];
38vector < pair<int,pair<int,int> > > edge[N];
39
40int s,t;
41
42void init()
43{
44 ms(inq,false);
45 ms(d,0x3f);
46 ms(p,0x3f);
47
48 for ( int i = 0 ; i <=n ; i++) edge[i].clear();
49}
50
51void spfa()
52{
53 queue<int>q;
54 q.push(s);
55 d[s] = 0 ;
56 p[s] = 0;
57 inq[s]=true;
58
59 while (!q.empty())
60 {
61 int now = q.front();
62// cout<<"now:"<<now<<endl;
63 q.pop();
64 inq[now] = false;
65
66 int siz = edge[now].size();
67 for ( int i = 0 ; i < siz ; i ++)
68 {
69 int v = edge[now][i].fst;
70 int nd = edge[now][i].sec.fst;
71 int nw = edge[now][i].sec.sec;
72 // cout<<"v:"<<v<<" nd:"<<nd<<" nw:"<<nw<<endl;
73 if (d[v]>d[now]+nd||((d[v]==d[now]+nd)&&(p[v]>p[now]+nw)))
74 {
75 d[v] = d[now] + nd;
76 p[v] = p[now] + nw;
77
78 if (inq[v]) continue;
79 q.push(v);
80 inq[v] = true;
81 }
82 }
83 }
84
85}
86int main()
87{
88 #ifndef ONLINE_JUDGE
89 freopen("code/in.txt","r",stdin);
90 #endif
91
92 while (scanf("%d%d",&n,&m)!=EOF)
93 {
94 if (n==0&&m==0) break;
95 init();
96 for ( int i = 1 ; i <= m ; i++)
97 {
98 int u,v,d,w;
99 scanf("%d%d%d%d",&u,&v,&d,&w);
100 edge[u].push_back(MP(v,MP(d,w)));
101 edge[v].push_back(MP(u,MP(d,w)));
102 }
103
104 scanf("%d%d",&s,&t);
105
106
107 spfa();
108
109 printf("%d %d\n",d[t],p[t]);
110 }
111
112 #ifndef ONLINE_JUDGE
113 fclose(stdin);
114 #endif
115 return 0;
116}