HUNNU 11568 Interstellar Travel(DP+优先队列)

Figure 1 Traveling through wormhole

The Hope‘s captain, Joseph Burnett, is working on a particular mission of geologic exploration of the Plant T. Now, his spaceship stays on Plant S. It has no energy and needs recharge. The Captain Burnett lists all wormholes that his spaceship can build between
all known plants. He needs to compute whether to transfer from these plant(s) and recharge there or build a wormhole from Plant S to Plant T directly. The reason is that the price of energy is different on different plants. For example, on plant DSB945, the
price of energy is 47 GUD (General Universe Dollar) per Bronto (a unit of energy) while on plant ZJT116, the price is only 16 GUD per Bronto. Given the price of every known plants, and list of wormholes that the spaceship can build, The Captain Burnett has
to plan the route in detail. He wants to travel to Plant T with minimum cost of money (General Universe Dollar).

Remember these two facts: the spaceship can’t recharge energy beyond its limit of maximum energy; if the energy of building a wormhole is beyond that the spaceship loads, the spaceship can’t build this wormhole.

The first line contains five integers: N (2 <= N <= 100) , M (1 <= M <= 200), L (1 <= L <= 100), S (1 <= S <= N), T (1 <= T <= N, T ≠ S). N is the number of known plants. M is the number of wormholes the spaceship can build. L is the limit of maximum energy,
which means the spaceship can load at most L Bronto energy. The Captain Burnett starts at Plant S and Plant T is his destination.

The second line contains N integers Pi (1 <= Pi <= 1000). They are price of energy in Plant 1~N.

M lines follows. Each line contains three integers: U (1 <= U <= N), V (1 <= V <= N, V ≠ U), E (1 <= E <= L). It means it will cost the spaceship E Bronto energy to build a wormhole from Plant U to Plant V or from Plant V to Plant U.

3 3 100 1 3
10 5 7
1 2 10
2 3 10
1 3 16
4 4 100 1 4
17 24 71 3
1 2 66
2 4 100
1 3 47
3 4 100
3 1 100 1 3
2 3 5
1 2 100

150
3284
Unreachable

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
const int N = 105 ;
const int INF = 1<<30;
struct NODE{
    int u,L,cost;
    friend bool operator<(NODE aa,NODE bb){
        return aa.cost>bb.cost;
    }
};
struct EDG{
    int to,next,L;
}edg[N<<3];
int eid , head[N] ;

void addEdg(int u,int v,int L){
    edg[eid].to=v; edg[eid].next=head[u] ;
    edg[eid].L=L; head[u]=eid++;

    edg[eid].to=u; edg[eid].next=head[v] ;
    edg[eid].L=L; head[v]=eid++;
}
int dp[N][N]  , p[N]; // dp[ 点 ] [ 剩于油量 ] ：最少花费
void init()
{
    eid=0;
    memset(head,-1,sizeof(head));
    for(int i=1; i<N; i++)
        for(int j=0; j<N; j++)
        dp[i][j]=INF;
}

int bfs(int vs , int vt , int n,int L)
{
    priority_queue<NODE>q;
    NODE now,pre;
    for(int i=1; i<=L; i++)
    {
        now.u=vs; now.cost=i*p[vs]; now.L=i;
        dp[vs][i]=now.cost;
        q.push(now);
    }
    while(!q.empty()){
        pre=q.top(); q.pop();
        if(dp[pre.u][pre.L]<pre.cost)continue;
        if(pre.u==vt) return pre.cost;
        for(int i=head[pre.u]; i!=-1; i=edg[i].next){
            if(pre.L<edg[i].L)continue;
            int v=edg[i].to;
            int lft=pre.L-edg[i].L;
            for(int j=lft; j<=L; j++ )
            if(dp[v][j]>pre.cost+(j-lft)*p[v]){
                dp[v][j] = pre.cost+(j-lft)*p[v];
                now.u=v; now.L=j; now.cost=dp[v][j];
                q.push(now);
            }
        }
    }
    return -1;
}
int main()
{
    int n,m,L,vs,vt , u , v, c;
    while(scanf("%d%d%d%d%d",&n,&m,&L,&vs,&vt)>0)
    {
        init();
        for(int i=1; i<=n; i++)
            scanf("%d",&p[i]);//在每个点补充的每升油的单价
        while(m--){
            scanf("%d%d%d",&u,&v,&c); //当前路（无向）的耗油量c
            addEdg(u , v , c);
        }
        int ans = bfs(vs , vt ,n ,L);
        if(ans!=-1)
            printf("%d\n",ans);
        else
            printf("Unreachable\n");
    }
}