UVA 10537 Toll! Revisited (逆推,最短路)

时间:2023-03-09 18:05:35
UVA 10537  Toll! Revisited (逆推,最短路)

从终点逆推,d[u]表示进入u以后剩下的货物,那么进入u之前的货物数量设为y,d[u] = x,那么y-x=ceil(y/20.0)=(y-1)/20+1=(y+19)/20。

(y-x)*20+r=y+19,0≤r≤19,即19*y=20*x+r,根据题意y应该尽量小,x的部分是不能变动的,所以y=x+ceil(x/19.0)。

然后从起点找一条字典序最小的路径即可,因为每个字母都是独一无二的,所以不必bfs,每次记录一个点就够了。

#include<bits/stdc++.h>
using namespace std; const int maxn = , maxm = maxn*maxn, town = , village = ;
bool g[maxn][maxn];
int head[maxn],nxt[maxm],to[maxm],ecnt;
int tp[maxn];
int id[];
char rid[]; void addEdge(int u,int v)
{
to[ecnt] = v;
nxt[ecnt] = head[u];
head[u] = ecnt++;
}
int id_cnt; void init()
{
memset(head,-,sizeof(head));
memset(id,-,sizeof(id));
memset(g,,sizeof(g));
ecnt = ;
id_cnt = ;
} inline int ID(char c) {
if(~id[c]) return id[c];
id[c]= id_cnt;
rid[id_cnt] = c;
tp[id_cnt] = 'a'<=c&&c<='z';//写成isalpha(c)-1;WA了
return id_cnt++;
}
typedef long long ll;
typedef pair<ll,int> Node;
#define fi first
#define se second ll d[maxn];
void dijkstra(int s,ll p)
{
memset(d,0x7f,sizeof(d));
priority_queue<Node,vector<Node>,greater<Node> > q;
q.push(Node(d[s] = p,s));
while(q.size()){
Node x = q.top(); q.pop();
int u = x.se;
if(d[u] != x.fi) continue;
ll t = (tp[u]?(+d[u]):((d[u]+)/+d[u]));
for(int i = head[u]; ~i; i = nxt[i]){
int v = to[i];
if(d[v] > t ){
q.push(Node(d[v]= t,v));
}
}
}
} ll cost(int u,int v)
{
return tp[v]?:((d[u]+)/);
} void FindPath(int s,int e)
{
int u = s;
while(u != e){
printf("%c-",rid[u]);
int nex = -;
for(int i = head[u]; ~i; i = nxt[i]){
int v = to[i];
if(d[u]- cost(u,v) >= d[v]){
if(~nex) {
if(rid[v] < rid[nex]) nex = v;
}else nex = v;
}
}
swap(nex,u);
}
printf("%c\n",rid[e]);
} int main()
{
//freopen("in.txt","r",stdin);
int n;
char a[],b[];
int kas = ;
while(scanf("%d",&n),~n){
init();
while(n--){
scanf("%s%s",a,b);
int u = ID(*a), v = ID(*b);
if(!g[u][v]){
g[u][v] = g[v][u] = true;
addEdge(u,v); addEdge(v,u);
}
}
ll p;
scanf("%lld%s%s",&p,a,b);
int s = ID(*a),e = ID(*b);
dijkstra(e,p);
printf("Case %d:\n%lld\n",++kas,d[s]);
FindPath(s,e);
}
return ;
}