洛谷 P3627 [APIO2009]抢掠计划 题解

Analysis

建图+强连通分量+SPFA求最长路

但要保证最后到达的点中包含酒馆

虽然思路并不难想，但要求的代码能力很高。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<queue>
 #define maxn 500010
 using namespace std;
 inline int read()
 {
     int x=;
     bool f=;
     char c=getchar();
     for(; !isdigit(c); c=getchar()) if(c=='-') f=;
     for(; isdigit(c); c=getchar()) x=(x<<)+(x<<)+c-'';
     if(f) return x;
     return -x;
 }
 inline void write(int x)
 {
     if(x<){putchar('-');x=-x;}
     if(x>)write(x/);
     putchar(x%+'');
 }
 struct node
 {
     int from,to,val,nex;
 }edge[*maxn];
 int n,m,s,p,cnt,ans,top,inl,num;
 int x[maxn],y[maxn],head[maxn],dfn[maxn],low[maxn],st[maxn],inn[maxn],sum[maxn],si[maxn],money[maxn],dis[maxn];
 bool book[maxn],book1[maxn],bar[maxn];
 inline void clear()
 {
     cnt=;
     memset(edge,,sizeof(edge));
     memset(head,,sizeof(head));
 }
 inline void add(int x,int y)
 {
     cnt++;
     edge[cnt].to=y;
     edge[cnt].nex=head[x];
     head[x]=cnt;
 }
 inline void build(int x,int y,int z)
 {
     cnt++;
     edge[cnt].from=x;
     edge[cnt].to=y;
     edge[cnt].val=z;
     edge[cnt].nex=head[x];
     head[x]=cnt;
 }
 inline void Tarjan(int u)
 {
     dfn[u]=low[u]=++num;
     st[++top]=u;
     for(int i=head[u];i;i=edge[i].nex)
     {
         int v=edge[i].to;
         if(!dfn[v])
         {
             Tarjan(v);
             low[u]=min(low[u],low[v]);
         }
         else if(!inn[v])
             low[u]=min(low[u],dfn[v]);
     }
     if(low[u]==dfn[u])
     {
         inn[u]=++inl;
         while(st[top]!=u)
         {
             if(bar[st[top]])book1[inl]=;
             sum[inl]+=money[st[top]];
             inn[st[top]]=inl;
              --top;
         }
         if(bar[st[top]])book1[inl]=;
         sum[inl]+=money[st[top]];
         --top;
     }
 }
 inline void spfa()
 {
     memset(book,,sizeof(book));
     for(int i=;i<=n;i++) dis[i]=;
     queue<int> q;
     q.push(inn[s]);
     book[inn[s]]=;
     dis[inn[s]]=sum[inn[s]];
     while(!q.empty())
     {
         for(int i=head[q.front()];i;i=edge[i].nex)
         {
             if(dis[edge[i].to]<dis[edge[i].from]+edge[i].val)
             {
                 dis[edge[i].to]=dis[edge[i].from]+edge[i].val;
                 if(book[edge[i].to]==)
                 {
                     q.push(edge[i].to);
                     book[edge[i].to]=;
                 }
             }
         }
         book[q.front()]=;
         q.pop();
     }
 }
 int main()
 {
     n=read();m=read();
     for(int i=;i<=m;i++)
     {
         x[i]=read();y[i]=read();
         add(x[i],y[i]);
     }
     for(int i=;i<=n;i++)money[i]=read();
     s=read();p=read();

     for(int i=;i<=p;i++)
     {
         int x;
         x=read();
         bar[x]=;
     }
     for(int i=;i<=n;i++)
         if(!dfn[i])
             Tarjan(i);

     clear();
     for(int i=;i<=m;i++)
         if(inn[x[i]]!=inn[y[i]])
             build(inn[x[i]],inn[y[i]],sum[inn[y[i]]]);
     spfa();
     for(int i=;i<=inl;i++)
         if(book1[i]==)
             ans=max(ans,dis[i]);
     write(ans);
     return ;
 }

请各位大佬斧正（反正我不认识斧正是什么意思）