【题解】：
最短路径问题，保证距离最短的同时，学妹权值最大，哈哈
【code】：
1 #include<iostream> 2 #include<queue> 3 #include<stdio.h> 4 #include<string.h> 5 #include<stdlib.h> 6 7 #define N 50005 8 #define INF 100000000 9 using namespace std; 10 11 struct Edge 12 { 13 int to; 14 int next; 15 int w; 16 int num; 17 }edge[N<<1]; 18 19 struct Nod 20 { 21 int u; 22 int dis; 23 int num; 24 }now,temp; 25 26 bool operator< (Nod a,Nod b) 27 { 28 if(a.dis!=b.dis) 29 return a.dis>b.dis; 30 return a.num<b.num; 31 } 32 33 int weight[N],head[N],visit[N]; 34 int dis[N]; 35 int ans[N]; 36 37 void init(int n) 38 { 39 int i; 40 for(i=0;i<=n;i++) 41 { 42 visit[i] = 0; 43 dis[i] = INF; 44 head[i] = -1; 45 ans[i]=-1; 46 } 47 } 48 49 void Dijkstra(int s) 50 { 51 int i,v; 52 dis[s] = 0; 53 ans[s] = weight[s]; 54 priority_queue<Nod> p_q; 55 temp.dis = 0; 56 temp.u = s; 57 temp.num = weight[s]; 58 p_q.push(temp); 59 while(!p_q.empty()) 60 { 61 temp = p_q.top(); 62 p_q.pop(); 63 if(visit[temp.u]) continue; 64 visit[temp.u] = 1; 65 for(i=head[temp.u];i!=-1;i=edge[i].next) 66 { 67 v = edge[i].to; 68 if(!visit[v]) 69 { 70 if(dis[v]>dis[temp.u]+edge[i].w||(dis[v]!=INF&&dis[v]==dis[temp.u]+edge[i].w&&ans[v]<ans[temp.u]+weight[v])) 71 { 72 dis[v] = dis[temp.u]+edge[i].w; 73 ans[v] = ans[temp.u]+weight[v]; 74 now.u = v; 75 now.dis = dis[v]; 76 now.num = ans[v]; 77 78 p_q.push(now); 79 } 80 } 81 } 82 } 83 } 84 85 int main() 86 { 87 int m,n; 88 while(~scanf("%d%d",&n,&m)) 89 { 90 int i; 91 for(i=0;i<n;i++) scanf("%d",weight+i); 92 int id = 0; 93 init(n); 94 int a,b,c; 95 for(i=0;i<m;i++) 96 { 97 scanf("%d%d%d",&a,&b,&c); 98 a--,b--; 99 edge[id].to = b; 100 edge[id].w = c; 101 edge[id].next = head[a]; 102 head[a] = id++; //将边 a --> b保存 103 104 edge[id].to = a; 105 edge[id].w = c; 106 edge[id].next = head[b]; 107 head[b] = id++; //将边 b --> a保存 108 } 109 Dijkstra(0); 110 printf("%d ",ans[n-1]); 111 } 112 return 0; 113 }