Problem Description
给你n个点，m条无向边，每条边都有长度d和花费p，给你起点s终点t，要求输出起点到终点的最短距离及其花费，如果最短距离有多条路线，则输出花费最少的。
Input
输入n,m，点的编号是1~n,然后是m行，每行4个数 a,b,d,p，表示a和b之间有一条边，且其长度为d，花费为p。最后一行是两个数 s,t;起点s，终点。n和m为0时输入结束。
(1<n<=1000, 0<m<100000, s != t)
(1<n<=1000, 0<m<100000, s != t)
Output
输出 一行有两个数， 最短距离及其花费。
Sample Input
3 2 1 2 5 6 2 3 4 5 1 3 0 0
Sample Output
9 11
Source
哎，终于出来了
1 #include<iostream>
2 #include<map>
3 #include<cstdio>
4 using namespace std;
5 #define inf 1000000000
6 int s, t;
7 int n, m,k;
8 struct di
9 {
10 int s, money;
11 }dis[1010];
12 struct e
13 {
14 int from, to, distance, cost;
15 }es[100010];
16 void ds()
17 {
18 dis[s].s = 0;
19 while (true)
20 {
21 bool update = false;
22 for (int i = 0; i < k; i++)
23 {
24 if (dis[es[i].from].s != inf && (dis[es[i].to].s > dis[es[i].from].s + es[i].distance ))
25 {
26 dis[es[i].to].s = dis[es[i].from].s + es[i].distance;
27 dis[es[i].to].money = dis[es[i].from].money + es[i].cost;
28 update = true;
29 }
30 else if(dis[es[i].to].s == dis[es[i].from].s + es[i].distance&&dis[es[i].to].money > dis[es[i].from].money + es[i].cost)
31 {
32 dis[es[i].to].s = dis[es[i].from].s + es[i].distance;
33 dis[es[i].to].money = dis[es[i].from].money + es[i].cost;
34 }
35 }
36 if (!update) break;
37 }
38 printf("%d %d\n", dis[t].s, dis[t].money);
39 }
40 int main()
41 {
42 while (cin >> n >> m, n != 0 && m != 0)
43 {
44 k = 0;
45 for (int i = 0; i < m; i++)
46 {
47 int a, b, c, d;
48 scanf("%d %d %d %d", &a,&b,&c,&d);
49 es[k].from = a, es[k].to = b, es[k].distance = c, es[k++].cost = d;
50 es[k].from = b, es[k].to = a, es[k].distance = c, es[k++].cost = d;
51 }
52 for (int i = 0; i <= n; i++)
53 {
54 dis[i].s = inf;
55 dis[i].money = 0;
56 }
57 scanf("%d %d", &s, &t);
58 ds();
59 }
60 return 0;
61 }
这是网上大佬的方法，为方便复习，拿来用了；
思路是djkstra的算法，当时我也想用了的，就是不太清楚cost给怎么定义，所以就用了bellman-ford，下次用队列优化的bellman-ford试试
1 #include <iostream>
2 #include <cstring>
3 #include <cstdio>
4 #define maxn 1007
5 #define INF 65535
6 using namespace std;
7
8 int start,e;
9 int n,m;
10 int map[maxn][maxn];
11 int cost[maxn][maxn];
12
13 void Dijkstra()
14 {
15 int v,Min,vis[maxn];
16 int d[maxn],c[maxn];
17 for(int i = 1;i <= n;i++) {
18 d[i] = map[start][i];
19 c[i] = cost[start][i];
20 }
21 memset(vis,0,sizeof(vis));
22 vis[start] = 1;
23 for(int i = 1;i <= n;i++)
24 {
25 if(vis[e]) break;
26 Min = INF;
27 for(int j = 1;j <= n;j++)
28 if(!vis[j] && d[j]<Min)
29 Min = d[v=j];
30 vis[v] = 1;
31 for(int j = 1;j <= n;j++)
32 if(!vis[j] && map[v][j]<INF) {
33 if(d[j] > d[v]+map[v][j]) {
34 d[j] = d[v]+map[v][j];
35 c[j] = c[v]+cost[v][j];
36 }
37 else if(d[j] == d[v]+map[v][j])
38 if(c[j] > c[v]+cost[v][j])
39 c[j] = c[v]+cost[v][j];
40 }
41 }
42 printf("%d %d\n",d[e],c[e]);
43 }
44
45 int main()
46 {
47 while(scanf("%d%d",&n,&m) && n+m)
48 {
49 for(int i = 1;i <= n;i++)
50 for(int j = 1;j <= n;j++) {
51 map[i][j] = i==j?0:INF;
52 cost[i][j] = i==j?0:INF;
53 }
54 int a,b,c,d;
55 for(int i = 1;i <= m;i++)
56 {
57 scanf("%d%d%d%d",&a,&b,&c,&d);
58 if(map[a][b]>c)
59 {
60 map[a][b]=map[b][a]=c;
61 cost[a][b]=cost[b][a]=d;
62 }
63 else if(map[a][b]==c)
64 {
65 if(cost[a][b]>d)
66 cost[a][b]=cost[b][a]=d;
67 }
68 }
69 scanf("%d%d",&start,&e);
70 Dijkstra();
71 }
72 return 0;
73 }