题目链接：http://poj.org/problem?id=3635
题目和最短路算法的形式很像，只要对每个状态确定转移的分支即可，其中每次油量只加一，因为后续一定会在队列中被取出来重新加一（如果可以的话）
代码：
#include<iostream> #include<cstring> #include<queue> using namespace std; #define maxn 1005 int head[maxn],nxt[maxn*20+10],ver[maxn*20+10],len[maxn*20+10]; int d[maxn][105]; int n,m,q; bool vis[maxn][105]; int price[maxn]; int tot=0; struct node{ int dist,city,fuel; bool operator < (const node &a)const{ return dist>a.dist; } node(){} node(int d,int c,int f):dist(d),city(c),fuel(f){} }; void addedge(int u,int v,int w){ ver[++tot]=v; len[tot]=w; nxt[tot]=head[u]; head[u]=tot; } int bfs(int start ,int end,int cab){ priority_queue<node> q; memset(d,0x3f,sizeof(d)); memset(vis,false,sizeof(vis)); q.push(node(0,start,0)); while(!q.empty()){ node cur=q.top(); q.pop(); if(cur.city == end)return cur.dist;//第一次出队一定是到达终点的最小花费 int city=cur.city,fuel=cur.fuel,dist=cur.dist; if(vis[city][fuel])continue; vis[city][fuel]=1; if(fuel+1<=cab && d[city][fuel+1]>dist+price[city]){ d[city][fuel+1]=dist+price[city]; q.push(node(d[city][fuel+1],city,fuel+1)); } for(int i=head[city];i;i=nxt[i]){ int to=ver[i],length=len[i]; if(length <= fuel){ if(d[to][fuel-length] > dist) { d[to][fuel-length]=dist; q.push(node(dist,to,fuel-length)); } } } } return -1; } int main(){ cin>>n>>m; for(int i=0;i<n;i++)scanf("%d",&price[i]); int u,v, w; for(int i=1;i<=m;i++){ scanf("%d%d%d",&u,&v,&w); addedge(u,v,w); addedge(v,u,w); } cin>>q; int st,ed,ca; while(q--){ scanf("%d%d%d",&ca,&st,&ed); int ans=bfs(st,ed,ca); if(ans==-1)cout<<"impossible"<<endl; else cout<<ans<<endl; } return 0; }