传送门
一道题让我又要学可行流又要学zkw费用流……
考虑一下,原题可以转化为一个有向图,每次走一条路径,把每一条边都至少覆盖一次,求最小代价
因为一条边每走过一次,就要付出一次代价
那不就是费用流了么
我们定义每走一次都会对一条边增加1的流量,1然后费用为时间
那么把下界设为1,上界设为inf,跑一个最小费用可行流就可以了
ps:不会可行流的可以去看看这个博客,我觉得写得很不错->这里
1 //minamoto 2 #include<iostream> 3 #include<cstdio> 4 #include<cstring> 5 #include<queue> 6 #define inf 0x3f3f3f3f 7 using namespace std; 8 #define getc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++) 9 char buf[1<<21],*p1=buf,*p2=buf;10 template<class T>inline bool cmax(T&a,const T&b){return a<b?a=b,1:0;}11 inline int read(){12 ????#define num ch-‘0‘13 ????char ch;bool flag=0;int res;14 ????while(!isdigit(ch=getc()))15 ????(ch==‘-‘)&&(flag=true);16 ????for(res=num;isdigit(ch=getc());res=res*10+num);17 ????(flag)&&(res=-res);18 ????#undef num19 ????return res;20 }21 const int N=505,M=100005;22 int head[N],Next[M],ver[M],edge[M],flow[M],tot=1;23 int dis[N],vis[N],S,T,ans;24 queue<int> q;25 inline void add(int u,int v,int e,int f){26 ????ver[++tot]=v,Next[tot]=head[u],head[u]=tot,edge[tot]=e,flow[tot]=f;27 ????ver[++tot]=u,Next[tot]=head[v],head[v]=tot,edge[tot]=-e,flow[tot]=0;28 }29 bool spfa(){30 ????memset(dis,-1,sizeof(dis));31 ????memset(vis,0,sizeof(vis));32 ????q.push(T),dis[T]=0,vis[T]=1;33 ????while(!q.empty()){34 ????????int u=q.front();q.pop();vis[u]=0;35 ????????for(int i=head[u];i;i=Next[i])36 ????????if(flow[i^1]){37 ????????????int v=ver[i],e=edge[i];38 ????????????if(dis[v]<0||dis[v]>dis[u]-e){39 ????????????????dis[v]=dis[u]-e;40 ????????????????if(!vis[v]) vis[v]=1,q.push(v);41 ????????????}42 ????????}43 ????}44 ????return ~dis[S];45 }46 int dfs(int u,int limit){47 ????if(!limit) return 0;48 ????if(u==T) return vis[T]=1,limit;49 ????int fl=0,f;vis[u]=1;50 ????for(int i=head[u];i;i=Next[i]){51 ????????int v=ver[i];52 ????????if(dis[v]==dis[u]-edge[i]&&!vis[v]&&(f=dfs(v,min(limit,flow[i])))){53 ????????????fl+=f,limit-=f;54 ????????????ans+=f*edge[i];55 ????????????flow[i]-=f,flow[i^1]+=f;56 ????????????if(!limit) break;57 ????????}58 ????}59 ????return fl;60 }61 void zkw(){62 ????while(spfa()){63 ????????vis[T]=1;64 ????????while(vis[T])65 ????????memset(vis,0,sizeof(vis)),dfs(S,inf);66 ????}67 }68 int d[N],n;69 int main(){70 ????//freopen("testdata.in","r",stdin);71 ????n=read();72 ????for(int i=1;i<=n;++i){73 ????????int t=read();74 ????????while(t--){75 ????????????int x=read(),y=read();76 ????????????--d[i],++d[x],ans+=y;77 ????????????add(i,x,y,inf);78 ????????}79 ????}80 ????S=0,T=n+2;81 ????for(int i=2;i<=n;++i) add(i,n+1,0,inf);82 ????for(int i=1;i<=n;++i){83 ????????if(d[i]>0)add(S,i,0,d[i]);84 ????????if(d[i]<0)add(i,T,0,-d[i]);85 ????}86 ????add(n+1,1,0,inf);87 ????zkw();88 ????printf("%d\n",ans);89 ????return 0;90 }
bzoj3876: [Ahoi2014&Jsoi2014]支线剧情(上下界费用流)
原文地址:https://www.cnblogs.com/bztMinamoto/p/9575318.html