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题目大意:每经过一个地方就要交出相应的货物作为过路费,问将一批货物从起点运到终点,最少需要携带多少货物?
题目分析:在每一站交的过路费由当前拥有的货物量来决定,所以,要以终点为源点,求一次单源最短路即可。注意,输出要求路径字典序最小。
代码如下:
# include<iostream> # include<cstdio> # include<queue> # include<vector> # include<cstring> # include<algorithm> using namespace std; # define LL long long const LL INF=0x7fffffffffffffff; struct Node { int u; LL d; Node(int _u,LL _d):u(_u),d(_d){} bool operator < (const Node &a) const { return d>a.d; } }; int vis[55],n,p,s,e,path[55]; LL dist[55]; vector<int>G[55]; LL get(LL d,int id) { if(id>25) return 1LL; LL l=d,r=d/(19LL)*(20LL)+19,ans; while(l<r){ LL m=l+(r-l)/2; LL k=m-(m+(LL)19)/((LL)20); if(k>=d){ r=m; ans=r; }else{ l=m+1; ans=l; } } return ans-d; /*LL temp=d*20LL/19; while(temp-(temp+19LL)/20LL<d) ++temp; return temp-d;*/ } void dijkstra() { for(int i=0;i<52;++i) path[i]=i; memset(vis,0,sizeof(vis)); priority_queue<Node>q; fill(dist,dist+52,INF); dist[e]=p; q.push(Node(e,(LL)p)); while(!q.empty()) { Node u=q.top(); q.pop(); if(vis[u.u]) continue; vis[u.u]=1; for(int i=0;i<G[u.u].size();++i){ int v=G[u.u][i]; LL w=get(u.d,u.u); if(dist[v]>w+u.d){ dist[v]=w+u.d; path[v]=u.u; q.push(Node(v,dist[v])); }else if(dist[v]==w+u.d&&path[v]>u.u){ path[v]=u.u; q.push(Node(v,dist[v])); } } } } void print(int u) { char c; if(u>25) c=u+‘a‘-26; else c=u+‘A‘; if(path[u]==u){ if(u!=s) printf("-"); printf("%c\n",c); return ; } if(u==s){ printf("%c",c); print(path[u]); }else{ printf("-%c",c); print(path[u]); } } int main() { int u,v,cas=0; char a[2],b[2],S[2],E[2]; while(scanf("%d",&n)&&n!=-1) { for(int i=0;i<52;++i) G[i].clear(); for(int i=0;i<n;++i){ scanf("%s%s",a,b); if(a[0]>=‘A‘&&a[0]<=‘Z‘) u=a[0]-‘A‘; else u=a[0]-‘a‘+26; if(b[0]>=‘A‘&&b[0]<=‘Z‘) v=b[0]-‘A‘; else v=b[0]-‘a‘+26; G[u].push_back(v); G[v].push_back(u); } scanf("%d%s%s",&p,S,E); if(S[0]>=‘A‘&&S[0]<=‘Z‘) s=S[0]-‘A‘; else s=S[0]-‘a‘+26; if(E[0]>=‘A‘&&E[0]<=‘Z‘) e=E[0]-‘A‘; else e=E[0]-‘a‘+26; dijkstra(); printf("Case %d:\n",++cas); printf("%lld\n",dist[s]); print(s); } return 0; }
UVA-10537 The Toll! Revisited (dijkstra)
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原文地址:http://www.cnblogs.com/20143605--pcx/p/4908664.html