这又是一道我没有当场想出来的题
题面
https://www.luogu.org/problemnew/show/P4016
题解
$S$连初始状态,末状态连$T$,转移边流量为$INF$,费用为$1$。
#include<cstdio> #include<cstring> #include<iostream> #include<vector> #include<queue> #define N 105 #define INF 1000000007 #define T (n+1) #define S 0 #define LL long long #define ri register int using namespace std; int n,a[N]; struct graph { vector<int> to,w,c; vector<int> ed[N]; LL dis[N]; int cur[N]; bool vis[N]; void add_edge(int a,int b,int aw,int ac) { to.push_back(b); w.push_back(aw); c.push_back(ac); ed[a].push_back(to.size()-1); to.push_back(a); w.push_back(0); c.push_back(-ac); ed[b].push_back(to.size()-1); } bool spfa() { memset(dis,0x2f,sizeof(dis)); memset(vis,0,sizeof(vis)); queue<int> q; dis[S]=0;q.push(S);vis[S]=1; while (!q.empty()) { int x=q.front(); q.pop(); for (ri i=0;i<ed[x].size();i++) { int e=ed[x][i]; if (dis[to[e]]>dis[x]+c[e] && w[e]) { dis[to[e]]=dis[x]+c[e]; if (!vis[to[e]]) vis[to[e]]=1,q.push(to[e]); } } vis[x]=0; } return dis[T]<INF; } int dfs(int x,int lim) { if (x==T || !lim) return lim; LL sum=0; vis[x]=1; for (ri &i=cur[x];i<ed[x].size();i++) { int e=ed[x][i]; if (dis[x]+c[e]==dis[to[e]] && w[e] && !vis[to[e]]) { int f=dfs(to[e],min(lim,w[e])); w[e]-=f; w[1^e]+=f; lim-=f; sum+=f; if (!lim) return sum; } } return sum; } LL zkw() { LL ret=0; while (spfa()) { memset(vis,0,sizeof(vis)); memset(cur,0,sizeof(cur)); ret+=dfs(S,INF)*dis[T]; } return ret; } } G; int suc(int x) { x++; if (x==n+1) return 1; else return x; } int pre(int x) { x--; if (x==0) return n; else return x; } int main() { scanf("%d",&n); int sum=0; for (ri i=1;i<=n;i++) scanf("%d",&a[i]),sum+=a[i]; int av=sum/n; for (ri i=1;i<=n;i++) { G.add_edge(S,i,a[i],0); G.add_edge(i,T,av,0); G.add_edge(i,suc(i),INF,1); G.add_edge(i,pre(i),INF,1); } cout<<G.zkw()<<endl; }