UVA 10537 The Toll! Revisited 最短路

d[u]表示从u离开时最少需要多少才能达到要求。

从终点开始往前更新，求出前继结点最少需要的d是多少

 

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<sstream>
#include<cmath>
#include<climits>
#include<string>
#include<map>
#include<queue>
#include<vector>
#include<stack>
#include<set>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
#define pb(a) push(a)
#define INF 0x1f1f1f1f
#define lson idx<<1,l,mid
#define rson idx<<1|1,mid+1,r
#define PI  3.1415926535898
template<class T> T min(const T& a,const T& b,const T& c) {
    return min(min(a,b),min(a,c));
}
template<class T> T max(const T& a,const T& b,const T& c) {
    return max(max(a,b),max(a,c));
}
void debug() {
#ifdef ONLINE_JUDGE
#else

    freopen("d:\\in1.txt","r",stdin);
    freopen("d:\\out1.txt","w",stdout);
#endif
}
int getch() {
    int ch;
    while((ch=getchar())!=EOF) {
        if(ch!=' '&&ch!='\n')return ch;
    }
    return EOF;
}

struct Edge
{
    int from,to,dist;
};
struct HeapNode
{
    ll d;
    int u;
    bool operator < (const HeapNode &ant ) const
    {
        return d>ant.d;
    }
};
ll BS(ll x,int u)
{
    if(u>=26)return x+1;
    ll l=0,r=LONG_LONG_MAX;
    while(l<r)
    {
        ll mid=(r+l)>>1;
        if(mid-(mid%20==0?mid/20:mid/20+1)>=x)
            r=mid;
        else l=mid+1;
    }
    return l;
}
const int maxn = 55;
const int n=52;
vector<int> g[maxn];
vector<Edge> edge;
int done[maxn];
ll d[maxn];
int p[maxn];
int toid(char c)
{
    if(c>='A'&&c<='Z')return c-'A';
    else return c-'a'+26;
}
char tochar(int id)
{
    if(id<26)return id+'A';
    else return id-26+'a';
}
void init()
{
    for(int i=0;i<n;i++)
        g[i].clear();
    edge.clear();
}
void add(int u,int v,int w)
{
    Edge e=(Edge){u,v,w};
    edge.push_back(e);
    g[u].push_back(edge.size()-1);
}
void dijksta(int e,int need)
{
    for(int i=0;i<n;i++)
        d[i]=LONG_LONG_MAX;
    memset(done,0,sizeof(done));
    priority_queue<HeapNode> q;

    d[e]=need;
    p[e]=-1;
    q.push((HeapNode){need,e});

    while(!q.empty())
    {
        HeapNode x=q.top();q.pop();
        int u=x.u;
        if(done[u])continue;
        done[u]=1;
        for(int i=0;i<g[u].size();i++)
        {
            int v=edge[g[u][i]].to;
            ll w=BS(d[u],u);
            if(w<d[v])
            {
                d[v]=w;
                p[v]=u;
                q.push((HeapNode){d[v],v});
            }
        }
    }
}
int main()
{
    int m;
    int ca=0;
    while(scanf("%d",&m)!=EOF&&m!=-1)
    {
        init();
        for(int i=1;i<=m;i++)
        {
            int u=toid(getch());
            int v=toid(getch());
            add(u,v,1);
            add(v,u,1);
        }
        int need;
        scanf("%d",&need);
        int s=toid(getch()),e=toid(getch());
        dijksta(e,need);
        ll res=d[s];
        printf("Case %d:\n",++ca);
        printf("%lld\n",res);

        int G[maxn][maxn]={0};
        for(int i=0;i<n;i++)
            for(int j=0;j<g[i].size();j++)
                G[i][edge[g[i][j]].to]=1;
        for(int u=s;;)
        {
            printf("%c%c",tochar(u),u==e?'\n':'-');
            if(u==e)break;
            for(int i=0;i<n;i++)
            {
                if(G[u][i]&&d[u]==BS(d[i],i))
                {
                    u=i;break;
                }
            }
        }
    }
    return 0;
}