UVa10537 Toll! Revisited

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题目描述 传送门

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考虑到反向做Dijkstra,终点的d值为p,反向Dijkstra时对于当前的(u,v)∈E可以推出如果v是村庄，边权为1；如果是城镇，边权为⌈d(u)/19⌉。为了方便，加了个虚拟终点0。

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代码

#include<cstdio>#include<iostream>#include<cstring>#include<algorithm>#include<queue>#include<cctype>#define g "%lld"using namespace std;typedef long long LL;const LL INF=1e18;const int maxn=256,maxm=(maxn-2)*(maxn-1);bool vis[maxn];struct Edge{    char t;    int nxt;    Edge(char a=0,int b=0):t(a),nxt(b){}}edges[maxm*2];struct heap{    LL d;    int v;    heap(LL a=0,int b=0):d(a),v(b){}    bool operator<(const heap&a)const{        return d>a.d;    }};char s[3],t[3];int h[maxn],n,m,cnt;LL d[maxn],p;void dfsput(char u){    if(u==0) return;    int nxt=-1;    LL minv=INF;    vis[u]=1;    for(int i=h[u];i>-1;i=edges[i].nxt) if(u==s[0]||(u!=s[0]&&d[edges[i].t]<d[u])){        if(d[edges[i].t]<minv||(d[edges[i].t]==minv&&edges[i].t<edges[nxt].t)) minv=d[edges[i].t],nxt=i;    }    printf("%c%c",u,edges[nxt].t==0?'\n':'-');    if(nxt>-1) dfsput(edges[nxt].t);}void addedge(int from,int to){    edges[++cnt]=Edge(to,h[from]);    h[from]=cnt;    edges[++cnt]=Edge(from,h[to]);    h[to]=cnt;}int main(){    int kase=0;    while(scanf("%d",&n)==1&&n!=-1){        printf("Case %d:\n",++kase);        cnt=-1;        memset(h,-1,sizeof(h));        for(int i=0;i<n;i++){            char c1[3],c2[3];            scanf("%s%s",c1,c2);            addedge(c1[0],c2[0]);        }               scanf("%lld%s%s",&p,s,t);        addedge(t[0],0);        memset(vis,0,sizeof(vis));        priority_queue<heap> q;        q.push(heap(p,0));        for(int i=1;i<=255;i++) d[i]=INF;        d[0]=p;        while(!q.empty()){            heap x=q.top();q.pop();            int u=x.v;            if(vis[u]) continue;            vis[u]=1;            for(int i=h[u];i>-1;i=edges[i].nxt){                int v=edges[i].t;                LL k;                if(v==s[0]) k=0;                else if(islower(v)) k=1;                else k=d[u]%19==0?d[u]/19:d[u]/19+1;                if(d[u]+k<d[v]){                    d[v]=d[u]+k;                    q.push(heap(d[v],v));                }            }        }        printf("%lld\n",d[s[0]]);        memset(vis,0,sizeof(0));        dfsput(s[0]);    }    return 0;}