UVA10537 The Toll! Revisited

Posted 2022-05-29 encodetalker

[vjudge]https://cn.vjudge.net/problem/UVA-10537

记\(dis_i\)为从\(i\)到\(ed\)最少需要多少单位的货物，这个东西可以直接dijkstra，初始条件\(dis_ed=x\)

输出方案的话直接从\(st\)开始找，每次找下一个点\(nxt\)时都要满足\(dis_now-cost_nxt==dis_nxt\)且\(nxt\)的字典序最小

唯一需要注意的是在做最短路的时候，由于我们这个过程是一个逆推的过程（即知道后继要求前面的点），这一部分的\(cost\)应该除以\(19\)而不是\(20\)

#include<iostream>
#include<string.h>
#include<string>
#include<stdio.h>
#include<algorithm>
#include<vector>
#include<math.h>
#include<queue>
#include<set>
#include<map>
using namespace std;
typedef long long ll;
typedef long double db;
const int N=10000;
const db pi=acos(-1.0);
#define lowbit(x) (x)&(-x)
#define sqr(x) (x)*(x)
#define rep(i,a,b) for (register int i=a;i<=b;i++)
#define per(i,a,b) for (register int i=a;i>=b;i--)
#define fir first
#define sec second
#define mp(a,b) make_pair(a,b)
#define pb(a) push_back(a)
#define maxd (ll)1e18
#define eps 1e-8
int n,m,mp[300];
ll dis[60],lim;
bool vis[60];
vector<int> sq[60];
char s[10],t[10];

int read()

    int x=0,f=1;char ch=getchar();
    while ((ch<'0') || (ch>'9')) if (ch=='-') f=-1;ch=getchar();
    while ((ch>='0') && (ch<='9')) x=x*10+(ch-'0');ch=getchar();
    return x*f;


void dij(int st,int ed,int lim)

    rep(i,1,n) dis[i]=maxd;
    dis[ed]=lim;
    memset(vis,0,sizeof(vis));
    rep(p,1,n)
    
        int best=0;ll mind=maxd;
        rep(i,1,n) 
            if ((!vis[i]) && (dis[i]<mind)) mind=dis[i];best=i;
        if (mind==maxd) return;
        vis[best]=1;
        int len=sq[best].size();
        rep(i,0,len-1)
        
            int v=sq[best][i];ll cost;
            if (best<27) cost=ceil((double)dis[best]/19.0);
            else cost=1;
            if ((dis[v]>dis[best]+cost) && (!vis[v]))
                dis[v]=dis[best]+cost;
        
    


int main()

    rep(i,1,26) mp['A'+i-1]=i;
    rep(i,1,26) mp['a'+i-1]=i+26;n=52;
    int cas=0;
    while ((scanf("%d",&m)==1) && (m!=-1))
    
        rep(i,1,n) sq[i].clear();
        rep(i,1,m)
        
            scanf("%s%s",s+1,t+1);
            int u=mp[s[1]],v=mp[t[1]];
            sq[u].pb(v);sq[v].pb(u);
        
        scanf("%lld%s%s",&lim,s+1,t+1);
        int st=mp[s[1]],ed=mp[t[1]];
        //cout << st << " " << ed << endl;
        dij(st,ed,lim);
        //rep(i,1,n) cout << dis[i] << " ";cout << endl;
        printf("Case %d:\n%lld\n",++cas,dis[st]);
        putchar(s[1]);
        int now=st;
        while (now!=ed)
        
            int len=sq[now].size(),best=100;
            rep(i,0,len-1)
            
                int v=sq[now][i],cost;
                if (v<27) cost=ceil((double)dis[now]/20);
                else cost=1;
                if ((dis[now]-cost==dis[v]) && (v<best)) best=v;
            
            now=best;putchar('-');
            if (now<27) putchar('A'+now-1);else putchar('a'+now-27);
        
        puts("");
    
    return 0;