题目连接：
http://codeforces.com/gym/100269/attachments
Description
Little Vasya is playing a new game named “Dwarf Tower”. In this game there are n different items,
which you can put on your dwarf character. Items are numbered from 1 to n. Vasya wants to get the
item with number 1.
There are two ways to obtain an item:
? You can buy an item. The i-th item costs ci money.
? You can craft an item. This game supports only m types of crafting. To craft an item, you give
two particular different items and get another one as a result.
Help Vasya to spend the least amount of money to get the item number 1.
Input
The first line of input contains two integers n and m (1 ≤ n ≤ 10 000; 0 ≤ m ≤ 100 000) — the number
of different items and the number of crafting types.
The second line contains n integers ci — values of the items (0 ≤ ci ≤ 109
).
The following m lines describe crafting types, each line contains three distinct integers ai, xi, yi — ai is the item that can be crafted from items xi and yi (1 ≤ ai , xi , yi ≤ n; ai ?= xi ; xi ?= yi ; yi ?= ai).
Output
The output should contain a single integer — the least amount of money to spend.
Sample Input
5 3
5 0 1 2 5
5 2 3
4 2 3
1 4 5
Sample Output
2
题意：
对与一个物品，你可以选择购买获得，但是要花费ci ， 或者是通过 xi yi 合成。
要你用最小的花费得到物品1.
题解：
我们对于每一个物品都应该花最小的花费的到。 a可以通过x， y合成。那么从x去到a的费用就是 c[y] 。（因为你已经跑到了x点，表示你已经有了x了）。
在建一个大原点 0， 0到每个点的费用为c[i] 。然后用0这里跑一次Dij 。这样你就可以得到了获得物品i 的最小花费。
在跑一下m次合成，找到最小的花费。
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <string>
5 #include <algorithm>
6 #include <cmath>
7 #include <vector>
8 #include <queue>
9 #include <map>
10 #include <stack>
11 #include <set>
12 using namespace std;
13 typedef long long LL;
14 typedef unsigned long long uLL;
15 #define ms(a, b) memset(a, b, sizeof(a))
16 #define rep(a, b) for(int a = 0;a<b;a++)
17 #define rep1(a, b) for(int a = 1;a<=b;a++)
18 #define pb push_back
19 #define mp make_pair
20 #define eps 0.0000000001
21 #define IOS ios::sync_with_stdio(0);cin.tie(0);
22 const LL INF = 0x3f3f3f3f3f3f3f3f;
23 const int inf = 0x3f3f3f3f;
24 const int mod = 1e9+7;
25 const int maxn = 10000+10;
26 int c[maxn];
27 struct qnode {
28 int v;
29 LL c;
30 qnode(int _v=0, LL _c =0):v(_v), c(_c) {}
31 bool operator < (const qnode &r) const {
32 return c > r.c;
33 }
34 };
35 struct Edge {
36 int v, cost;
37 Edge(int _v=0, int _cost =0):v(_v), cost(_cost) {}
38 };
39 vector <Edge> E[10*maxn];
40 bool vis[maxn];
41 LL dist[maxn];
42 void Dij(int n, int start) {
43 memset(vis,false,sizeof(vis));
44 for(int i=1; i<=n; i++)dist[i]=INF;
45 priority_queue<qnode>que;
46 while(!que.empty())que.pop();
47 dist[start]=0;
48 que.push(qnode(start,0));
49 qnode tmp;
50 while(!que.empty()) {
51 tmp=que.top();
52 que.pop();
53 int u=tmp.v;
54 if(vis[u])continue;
55 vis[u]=true;
56 for(int i=0; i<E[u].size(); i++) {
57 int v=E[tmp.v][i].v;
58 int cost=E[u][i].cost;
59 if(!vis[v]&&dist[v]>dist[u]+cost) {
60 dist[v]=dist[u]+cost;
61 que.push(qnode(v,dist[v]));
62 }
63 }
64 }
65 }
66 void addedge(int u, int v, int w) {
67 E[u].pb(Edge(v, w));
68 }
69 vector<pair<int, int> > One;
70 void solve() {
71 int n, m, x, a, b;
72 scanf("%d%d", &n, &m);
73 for(int i = 1; i<=n; i++) {
74 scanf("%d", &c[i]);
75 addedge(0, i, c[i]);//0指向物品i，表示直接购买的花费
76 }
77 for(int i = 1; i<=m; i++) {
78 scanf("%d%d%d", &x, &a, &b);
79 addedge(a, x, c[b]);//a指向物品x，表示需要在花费c[b]的花费就可以合成x
80 addedge(b, x, c[a]);
81 if(x==1){//记录一下物品1的合成
82 One.pb(mp(a, b));
83 }
84 }
85 Dij(n, 0);
86 LL ans = dist[1];//得到1的花费
87 for(int i = 0;i<One.size();i++){
88 ans = min(ans, dist[One[i].first]+dist[One[i].second]);//和通过合成的花费比较
89 }
90 printf("%lld\n", ans);
91 }
92 int main() {
93 #ifdef LOCAL
94 freopen("input.txt", "r", stdin);
95 // freopen("output.txt", "w", stdout);
96 #endif
97 freopen("dwarf.in", "r", stdin);
98 freopen("dwarf.out", "w", stdout);
99 solve();
100 return 0;
101 }