Codeforces Gym 100269 Dwarf Tower （最短路）

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。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <string>
 #include <algorithm>
 #include <cmath>
 #include <vector>
 #include <queue>
 #include <map>
 #include <stack>
 #include <set>
 using namespace std;
 typedef long long LL;
 typedef unsigned long long uLL;
 #define ms(a, b) memset(a, b, sizeof(a))
 #define rep(a, b) for(int a = 0;a<b;a++)
 #define rep1(a, b) for(int a = 1;a<=b;a++)
 #define pb push_back
 #define mp make_pair
 #define eps 0.0000000001
 #define IOS ios::sync_with_stdio(0);cin.tie(0);
 const LL INF = 0x3f3f3f3f3f3f3f3f;
 const int inf = 0x3f3f3f3f;
 const int mod = 1e9+;
 const int maxn = +;
 int c[maxn];
 struct qnode {
     int v;
     LL c;
     qnode(int _v=, LL _c =):v(_v), c(_c) {}
     bool operator < (const qnode &r) const {
         return c > r.c;
     }
 };
 struct Edge {
     int v, cost;
     Edge(int _v=, int _cost =):v(_v), cost(_cost) {}
 };
 vector <Edge> E[*maxn];
 bool vis[maxn];
 LL dist[maxn];
 void Dij(int n, int start) {
     memset(vis,false,sizeof(vis));
     for(int i=; i<=n; i++)dist[i]=INF;
     priority_queue<qnode>que;
     while(!que.empty())que.pop();
     dist[start]=;
     que.push(qnode(start,));
     qnode tmp;
     while(!que.empty()) {
         tmp=que.top();
         que.pop();
         int u=tmp.v;
         if(vis[u])continue;
         vis[u]=true;
         for(int i=; i<E[u].size(); i++) {
             int v=E[tmp.v][i].v;
             int cost=E[u][i].cost;
             if(!vis[v]&&dist[v]>dist[u]+cost) {
                 dist[v]=dist[u]+cost;
                 que.push(qnode(v,dist[v]));
             }
         }
     }
 }
 void addedge(int u, int v, int w) {
     E[u].pb(Edge(v, w));
 }
 vector<pair<int, int> > One;
 void solve() {
     int n, m, x, a, b;
     scanf("%d%d", &n, &m);
     for(int i = ; i<=n; i++) {
         scanf("%d", &c[i]);
         addedge(, i, c[i]);//0指向物品i，表示直接购买的花费
     }
     for(int i = ; i<=m; i++) {
         scanf("%d%d%d", &x, &a, &b);
         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
         addedge(b, x, c[a]);
         if(x==){//记录一下物品1的合成
             One.pb(mp(a, b));
         }
     }
     Dij(n, );
     LL ans = dist[];//得到1的花费
     for(int i = ;i<One.size();i++){
         ans = min(ans, dist[One[i].first]+dist[One[i].second]);//和通过合成的花费比较
     }
     printf("%lld\n", ans);
 }
 int main() {
 #ifdef LOCAL
     freopen("input.txt", "r", stdin);
 //        freopen("output.txt", "w", stdout);
 #endif
     freopen("dwarf.in", "r", stdin);
     freopen("dwarf.out", "w", stdout);
     solve();
     return ;
 }