UVA11367 Full Tank?

题面传送门
二维$dj$
考虑设$d_{i,j}$为到达$i$点，还有$j$升油的最小花费。
那么可以拓展出两种状态，即加$1$升油或前往下一个城市。
代码实现:

#include
#include
#include
using namespace std;
int n,m,k,x,y,z,d[1039][139],w[1039],cur;
struct yyy{
     int to,w,z;}tmp;
struct ljb{
     
	int head,h[1039];
	yyy f[20039];
	inline void add(int x,int y,int z){
     
		f[++head]=(yyy){
     y,z,h[x]};
		h[x]=head;
	}
}s;
struct astar{
     
    int to,yo,mo;
    bool operator<(const astar &x) const {
     
        return mo>x.mo;
    }
};
priority_queue<astar>q;
int main(){
     
//	freopen("1.in","r",stdin);
//	freopen("1.out","w",stdout);
	memset(s.h,-1,sizeof(s.h));
	register int i;
	register astar now;
	scanf("%d%d",&n,&m);
	for(i=1;i<=n;i++) scanf("%d",&w[i]);
	for(i=1;i<=m;i++){
     
		scanf("%d%d%d",&x,&y,&z);
		s.add(x+1,y+1,z);s.add(y+1,x+1,z);
	}
	scanf("%d",&k);
	for(i=1;i<=k;i++){
     
		scanf("%d%d%d",&z,&x,&y);
		x++;y++;
		memset(d,0x3f,sizeof(d));
		while(!q.empty()) q.pop();
		d[x][0]=0;q.push((astar){
     x,0,0});
		while(!q.empty()){
     
			now=q.top();
			if(now.to==y) break;
			q.pop();
			cur=s.h[now.to];
			if(now.yo!=z&&d[now.to][now.yo+1]>now.mo+w[now.to]) d[now.to][now.yo+1]=now.mo+w[now.to],q.push((astar){
     now.to,now.yo+1,now.mo+w[now.to]});
			while(cur!=-1){
     
				tmp=s.f[cur];
				if(now.yo>=tmp.w&&d[tmp.to][now.yo-tmp.w]>now.mo)d[tmp.to][now.yo-tmp.w]=now.mo,q.push((astar){
     tmp.to,now.yo-tmp.w,now.mo});
				cur=tmp.z;
			}
		}
		if(now.to!=y) printf("impossible\n");
		else printf("%d\n",now.mo);
	}
}