UVA- 10841 - Lift Hopping in the Real World(dijkstra)

题意:电梯调度的加强版,有N个电梯,每个电梯只能到规定的楼层停,从A电梯到B电梯时需要花5秒时间,并且还需要等待B电梯停到这层花费的时间(按最糟糕的情况处理)

 分析：因为最优策略是每个电梯最多搭乘一次，若搭乘多次则花费的时间会更多..所以不用考虑每个电梯搭乘后的状态..

       松弛时一定是一个电梯到另一个电梯,因为在同一个电梯无法松弛!

// File Name: 10841.cpp
// Author: Zlbing
// Created Time: 2013/5/19 12:58:02

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<set>
#include<map>
#include<vector>
#include<cstring>
#include<stack>
#include<cmath>
#include<queue>
using namespace std;
#define CL(x,v); memset(x,v,sizeof(x));
#define INF 0x3f3f3f3f
#define LL long long
#define REP(i,r,n) for(int i=r;i<=n;i++)
#define RREP(i,n,r) for(int i=n;i>=r;i--)
const int MAXN=150;

int n,k;
int T[MAXN];
int G[MAXN][MAXN];
int f[MAXN];
int maxn[MAXN];
int minn[MAXN];
struct node{
    int u,cost;
    bool operator <(const node& rhs)const{
        return cost>rhs.cost;
    }
};
int vis[MAXN];
int d[MAXN];
int dij()
{
    CL(vis,0);
    CL(d,-1);
    priority_queue<node> Q;
    Q.push((node){0,0});
    node t,tt;
    while(!Q.empty())
    {
        t=Q.top();
        Q.pop();
        if(t.u==k)return t.cost;
        int u=t.u;
        int cost=t.cost;
        if(vis[u])continue;
        vis[u]=1;
        for(int i=0;i<n;i++)
            if(G[i][u]){
                int a=T[i]*(maxn[i]-u);
                int b=T[i]*(u-minn[i]);
                int tmp=max(a,b);
                for(int j=0;j<105;j++)
                {
                    if(u==j)continue;
                    if(G[i][j])
                    {
                        if(u==0)
                        {
                            tt.u=j;
                            tt.cost=tmp+cost+abs(j-u)*T[i];
                            Q.push(tt);
                            //printf("i=%d %d->%d cost=%d\n",i,u,tt.u,tt.cost);
                        }
                        else{
                            tt.u=j;
                            tt.cost=tmp+cost+abs(j-u)*T[i]+5;
                            Q.push(tt);
                            //printf("i=%d %d->%d cost=%d\n",i,u,tt.u,tt.cost);
                        }
                    }
                }
            }
    }
    return -1;
}
int main()
{
    while(~scanf("%d%d",&n,&k))
    {
        REP(i,0,n-1)
        {
            scanf("%d",&T[i]);
        }
        CL(G,0);
        CL(f,0);
        char ch[MAXN*10];
        CL(minn,-1);
        CL(maxn,-1);
        REP(i,0,n-1)
        {
            int a=0;
            do{
                scanf("%d",&a);
                G[i][a]=1;
                if(maxn[i]==-1)
                    maxn[i]=a;
                else maxn[i]=max(maxn[i],a);
                if(minn[i]==-1)
                    minn[i]=a;
                else minn[i]=min(minn[i],a);
            }while(getchar()!='\n');
        }
        int ans=dij();
        if(ans==-1)
            printf("IMPOSSIBLE\n");
        else printf("%d\n",ans);
    }
    return 0;
}