BZOJ 2324 营救皮卡丘

http://www.lydsy.com/JudgeOnline/problem.php?id=2324

思路:最小费用最大流

考虑设数组d[k][i][j]，代表只用前k个城市，i到j的最短路

然后可以这样建图

S->0 流量为K，费用为0

i->i+n 流量为inf，费用为0

i+n->T 流量为1

i+n->j 流量为inf，费用为d[j][i][j]

S->i  流量为1，费用为0 代表如果一个人到了i城市，它可以继续走，从而续出来的一个新流量。

 #include<algorithm>
 #include<cstdio>
 #include<cmath>
 #include<cstring>
 #include<iostream>
 #define inf 0x7fffffff
 int d[][][],a[][],K;
 int tot,go[],next[],first[],cost[],flow[];
 int op[],dis[],c[],vis[],edge[],from[];
 int S,T,n,m,ans;
 int read(){
     char ch=getchar();int t=,f=;
     while (ch<''||ch>''){if (ch=='-') f=-;ch=getchar();}
     while (''<=ch&&ch<=''){t=t*+ch-'';ch=getchar();}
     return t*f;
 }
 void insert(int x,int y,int z,int l){
     tot++;
     go[tot]=y;
     next[tot]=first[x];
     first[x]=tot;
     flow[tot]=z;
     cost[tot]=l;
 }
 void add(int x,int y,int z,int l){
     insert(x,y,z,l);op[tot]=tot+;
     insert(y,x,,-l);op[tot]=tot-;
 }
 bool spfa(){
     for (int i=;i<=T;i++)
      dis[i]=0x3f3f3f3f,vis[i]=;
     int h=,t=;c[]=S;vis[S]=;dis[S]=;
     while (h<=t){
         int now=c[h++];
         for (int i=first[now];i;i=next[i]){
             int pur=go[i];
             if (dis[pur]>dis[now]+cost[i]&&flow[i]){
                 dis[pur]=dis[now]+cost[i];
                 edge[pur]=i;
                 from[pur]=now;
                 if (vis[pur]) continue;
                 vis[pur]=;
                 c[++t]=pur;
             }
         }
         vis[now]=;
     }
     return dis[T]!=0x3f3f3f3f;
 }
 void updata(){
     int mn=0x7ffffff;
     for (int i=T;i!=S;i=from[i]){
         mn=std::min(mn,flow[edge[i]]);
     }
     for (int i=T;i!=S;i=from[i]){
         ans+=mn*cost[edge[i]];
         flow[edge[i]]-=mn;
         flow[op[edge[i]]]+=mn;
     }
 }
 int main(){
     n=read();m=read();K=read();
     for (int i=;i<=n;i++)
      for (int j=;j<=n;j++)
       if (i!=j){
             a[i][j]=0x3f3f3f3f;
       }else
             a[i][j]=;
     for (int i=;i<=m;i++){
         int u=read(),v=read(),w=read();
         a[u][v]=std::min(a[u][v],w);
         a[v][u]=std::min(a[v][u],w);
     }
     for (int k=;k<=n;k++){
         for (int i=;i<=n;i++)
          for (int j=;j<=n;j++)
           a[i][j]=std::min(a[i][k]+a[k][j],a[i][j]);
         for (int i=;i<=n;i++)
          for (int j=;j<=n;j++)
           d[k][i][j]=a[i][j];
     }
     S=*n+;T=S+;
     add(S,,K,);
     for (int i=;i<=n;i++)
      add(i,i+n+,inf,);
     for (int i=;i<=n;i++)
      add(i,T,,);
     for (int i=;i<=n;i++)
      add(S,i+n+,,);
     for (int i=;i<=n;i++)
      for (int j=i+;j<=n;j++)
       if (d[j][i][j]<0x3f3f3f3f)
       add(i+n+,j,inf,d[j][i][j]);
     while (spfa()) updata();
     printf("%d\n",ans);
 }