《算法竞赛进阶指南》0x26广搜变形 POJ3635

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;
}