以状态（u,fuel)为结点建图（把每个点拆成100个），表示在点u时还剩下fuel个单位的燃料，那么状态就可以这样转移：
(u,fuel)->(u,fuel+i) ： 加i单位的燃料，所以这条边的权值就是 i*p[u];
(u,fuel) ->(v,fuel-dist): 走到另一个点，其中dist为路径。
当然直接这样做可能会超时，需要做两个小优化（做其中一个就能AC了）
1. (u,fuel)->(u,fuel+i) 中的i只需为1就行了，这样就能边数从100^2降到100. 比如(u,fuel)->(u,fuel+3)可以用
(u,fuel)->(u,fuel+1)->(u,fuel+2)->(u,fuel+3) 代替
2. 没必要计算全部结点，只要确定了d[u][fuel]的值 就break;
//#pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<iostream> #include<sstream> #include<cmath> #include<climits> #include<string> #include<map> #include<queue> #include<vector> #include<stack> #include<set> using namespace std; typedef long long ll; typedef unsigned long long ull; typedef pair<int,int> pii; #define pb(a) push(a) #define INF 0x1f1f1f1f #define lson idx<<1,l,mid #define rson idx<<1|1,mid+1,r #define PI 3.1415926535898 template<class T> T min(const T& a,const T& b,const T& c) { return min(min(a,b),min(a,c)); } template<class T> T max(const T& a,const T& b,const T& c) { return max(max(a,b),max(a,c)); } void debug() { #ifdef ONLINE_JUDGE #else freopen("in.txt","r",stdin); //freopen("d:\out1.txt","w",stdout); #endif } int getch() { int ch; while((ch=getchar())!=EOF) { if(ch!=' '&&ch!=' ')return ch; } return EOF; } struct HeapNode { int d,u,fuel; bool operator <(const HeapNode &ant) const { return ant.d<d; } }; struct Edge { int u,v,w; }; const int maxn=1005; const int maxc=105; vector<int> g[maxn]; vector<Edge> edge; int p[maxn]; int n; void init() { for(int i=0;i<n;i++) g[i].clear(); edge.clear(); } void add(int u,int v,int w) { Edge e=(Edge){u,v,w}; edge.push_back(e); g[u].push_back(edge.size()-1); } bool done[maxn][maxc]; int d[maxn][maxc]; void solve(int s,int e,int c) { memset(done,0,sizeof(done)); memset(d,INF,sizeof(d)); d[s][0]=0; priority_queue<HeapNode> q; q.push((HeapNode){0,s,0}); while(!q.empty()) { HeapNode x=q.top();q.pop(); if(done[x.u][x.fuel]) continue; int u=x.u,fuel=x.fuel; done[u][fuel]=1; if(u==e&&fuel==0)break; if(fuel!=c&&d[u][fuel]+p[u]<d[u][fuel+1]) { d[u][fuel+1]=d[u][fuel]+p[u]; q.push((HeapNode){d[u][fuel+1],u,fuel+1}); } for(int i=0;i<g[u].size();i++) { int v=edge[g[u][i]].v; int w=edge[g[u][i]].w; if(fuel>=w&&d[u][fuel]<d[v][fuel-w]) { d[v][fuel-w]=d[u][fuel]; q.push((HeapNode){d[v][fuel-w],v,fuel-w}); } } } if(d[e][0]!=INF) printf("%d ",d[e][0]); else printf("impossible "); } int main() { int m; while(scanf("%d%d",&n,&m)!=EOF) { init(); for(int i=0;i<n;i++) scanf("%d",&p[i]); for(int i=0;i<m;i++) { int u,v,w; scanf("%d%d%d",&u,&v,&w); add(u,v,w); add(v,u,w); } int q; scanf("%d",&q); for(int i=0;i<q;i++) { int s,e,c; scanf("%d%d%d",&c,&s,&e); solve(s,e,c); } } return 0; }