Face questions
https://www.luogu.org/problem/P4307
answer
Japanese film out of $ aysn $.
Points to $ 2n + m $ bipartite graph approach can easily occur.
The point everyone split into success and failure points, then a person is to win a few games and all the people of his game and the number of field failures, limit traffic in the new point on the line.
More preferred approach, points, $ n + m $:
Adjustment Act.
Assume that everyone is lost, and then a game, win the person is wins +1, -1 defeat games, lose unchanged.
Such a seemingly put into a binary function of the membership function.
Then count the right side, according to a method of connecting directly above the edge on the line, without dismantling point.
#include<queue> #include<vector> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define N 10050 #define INF 1000000007 #define S 0 #define T (n+m+1) #define LL long long #define ri register int using namespace std; int n,m; int a[N],b[N],preb[N],c[N],d[N],e[N]; int u[N],v[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,0x3f,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 main(){ scanf("%d %d",&n,&m); for (ri i=1;i<=n;i++) { scanf("%d %d %d %d",&a[i],&b[i],&c[i],&d[i]); preb[i]=b[i]; } for (ri i=1;i<=m;i++) { scanf("%d %d",&u[i],&v[i]); b[u[i]]++; b[v[i]]++; } for (ri i=1;i<=n;i++) e[i]=a[i]+b[i]; for (ri i=1;i<=m;i++) { G.add_edge(S,i,1,0); G.add_edge(i,m+u[i],1,0); G.add_edge(i,m+v[i],1,0); } for (ri i=1;i<=n;i++) { for (ri j=a[i];j+preb[i]<e[i];j++) G.add_edge(m+i,T,1,2*j*c[i]+c[i]-2*d[i]*e[i]+2*j*d[i]+d[i]); } LL ans=0; for (ri i=1;i<=n;i++) ans+=b[i]*b[i]*d[i]+a[i]*a[i]*c[i]; ans+=G.zkw(); cout<<ans<<endl; }