UVA 10537 Toll! Revisited

这个题写的。。有点纠结啊。主要是如果从某地运东西到一个城镇时，根据已知城镇的spoons的数量倒推回去求原来的那个地点spoons的数量有点点麻烦，这里处理好了就应该没有问题了。还有这个字典序的问题。。这里ASCII码越小，字典序越小。

#include <iostream>
#include <cstring>
#include <queue>
#include <vector>
#include <cstdio>
#define INF 0x3f3f3f3f3f3f3f3fLL
#define maxn 60
using namespace std;
typedef pair <long long int,int> pii;
long long int d[maxn],p;
int first[maxn];
vector<int> st,v,next,w,path;
int n,m,e,start,end;

void init()
{
    e = 0;
    st.clear();
    v.clear();
    next.clear();
    w.clear();
    path.clear();
    memset(first,-1,sizeof(first));
}
void add_edge(int a,int b)
{
    //st[e] = a;
    st.push_back(a);
    //v[e] = b;
    v.push_back(b);
    //next[e] = first[a];
    next.push_back(first[a]);
    w.push_back(0);
    first[a] = e;
    e++;
}
void Read_Graph()
{
    int i,a,b;
    char ch1,ch2;
    e = 0;
    getchar();
    for(i = 1;i <= m;i++)
    {
        scanf("%c %c",&ch1,&ch2);
        getchar();
        a = ch1 - 'A';
        b = ch2 - 'A';
        add_edge(a,b);
        add_edge(b,a);
    }
    scanf("%lld %c %c",&p,&ch1,&ch2);
    start = ch1 - 'A';
    end = ch2 - 'A';
}

void dijkstra()
{
    priority_queue < pii,vector<pii>,greater<pii> > q;

    int i;
    for(i = 0;i < maxn;i++)
    {
        d[i] = INF;
    }
    d[end] = p;
    q.push(make_pair(p,end));
    while(!q.empty())
    {
        while(!q.empty() && q.top().first > d[q.top().second])   q.pop();
        int u = q.top().second;
        q.pop();
        if(u == start)  break;
        for(i = first[u];i != -1;i = next[i])
        {
            if(u < 26)
            {
                if(d[u] % 19 == 0)
                {
                    if(d[v[i]] > d[u] / 19 * 20)
                    {
                        d[v[i]] = d[u] / 19 * 20;
                        w[i] = w[i ^ 1] = d[v[i]] - d[u];
                        q.push(make_pair(d[v[i]],v[i]));
                    }
                }
                else if((d[u] + 1) % 19)
                {
                    if(d[v[i]] > (d[u] + 1) * 20 / 19)
                    {
                        d[v[i]] = (d[u] + 1) * 20 / 19;
                        w[i] = w[i ^ 1] = d[v[i]] - d[u];
                        q.push(make_pair(d[v[i]],v[i]));
                    }
                }
                else
                {
                    if(d[v[i]] > (d[u] + 1) * 20 / 19 - 1)
                    {
                        d[v[i]] = (d[u] + 1) * 20 / 19 - 1;
                        w[i] = w[i ^ 1] = d[v[i]] - d[u];
                        q.push(make_pair(d[v[i]],v[i]));
                    }
                }
            }
            else
            {
                if(d[v[i]] > d[u] + 1)
                {
                    d[v[i]] = d[u] + 1;
                    w[i] = w[i ^ 1] = 1;
                    q.push(make_pair(d[v[i]],v[i]));
                }
            }
        }
    }
}

void solve()
{
    int i,temp,now = start;
    path.push_back(now);
    while(now != end)
    {
        temp = 1 << 6;
        for(i = 0;i < e;i++)
        {
            if(d[now] == d[v[i]] + w[i] && st[i] == now && w[i] != 0 && v[i] < temp)
            {
                temp = v[i];
            }
        }
        now = temp;
        path.push_back(now);
    }
    printf("%lld\n",d[start]);
    printf("%c",path[0] + 'A');
    for(i = 1;i < path.size();i++)
        printf("-%c",path[i] + 'A');
    printf("\n");
}

int main()
{
    int ncase = 0;
    while(scanf("%d",&m))
    {
        if(m == -1) break;
        init();
        Read_Graph();
        dijkstra();
        printf("Case %d:\n",++ncase);
        solve();
    }
    return 0;
}