BZOJ 2039 人员雇佣(最小割)
最小割的建图模式一般是,先算出总收益,然后再通过网络模型进行割边减去部分权值。
然后我们需要思考什么才能带来收益,什么才能有权值冲突。
s连向选的点,t连向不选的点,那么收益的减少量应该就是将s集和t集分开的割边集。
下面说这道题的建图:
点:
每个人一个点,额外设源汇点。
边:
源向人连这个人能造成的全部收益(当作雇佣所有人,然后此人造成的收益)
人与人之间连两人熟悉度*2,呃,题意问题。
人向汇连雇佣需要花的钱。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
# include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # include <vector> # include <queue> # include <stack> # include <map> # include <bitset> # include <set> # include <cmath> # include <algorithm> using namespace std; # define lowbit(x) ((x)&(-x)) # define pi acos(-1.0) # define eps 1e-8 # define MOD 1000000007 # define INF (LL)1<<60 # define mem(a,b) memset(a,b,sizeof(a)) # define FOR(i,a,n) for(int i=a; i<=n; ++i) # define FO(i,a,n) for(int i=a; i<n; ++i) # define bug puts("H"); # define lch p<<1,l,mid # define rch p<<1|1,mid+1,r # define mp make_pair # define pb push_back typedef pair<int,int> PII; typedef vector<int> VI; # pragma comment(linker, "/STACK:1024000000,1024000000") typedef long long LL; int Scan() { int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } const int N=1005; //Code begin... struct Edge{int p, next; LL w;}edge[N*N*5]; int head[N], cnt=2, s, t, vis[N]; queue<int>Q; LL ss[N]; void add_edge(int u, int v, LL w){ edge[cnt].p=v; edge[cnt].w=w; edge[cnt].next=head[u]; head[u]=cnt++; edge[cnt].p=u; edge[cnt].w=0; edge[cnt].next=head[v]; head[v]=cnt++; } int bfs(){ int i, v; mem(vis,-1); vis[s]=0; Q.push(s); while (!Q.empty()) { v=Q.front(); Q.pop(); for (int i=head[v]; i; i=edge[i].next) { if (edge[i].w>0&&vis[edge[i].p]==-1) vis[edge[i].p]=vis[v]+1, Q.push(edge[i].p); } } return vis[t]!=-1; } LL dfs(int x, LL low){ int i; LL a, temp=low; if (x==t) return low; for (int i=head[x]; i; i=edge[i].next) { if (edge[i].w>0&&vis[edge[i].p]==vis[x]+1) { a=dfs(edge[i].p,min(edge[i].w,temp)); temp-=a; edge[i].w-=a; edge[i^1].w+=a; if (temp==0) break; } } if (temp==low) vis[x]=-1; return low-temp; } LL dinic(){ LL sum=0; while (bfs()) sum+=dfs(s,INF); return sum; } int main () { int n; LL ans=0, x; scanf("%d",&n); s=0; t=n+1; FOR(i,1,n) scanf("%lld",&x), add_edge(i,t,x); FOR(i,1,n) FOR(j,1,n) { scanf("%lld",&x); ss[i]+=x; if (i==j||!x) continue; add_edge(i,j,x*2); } FOR(i,1,n) add_edge(s,i,ss[i]), ans+=ss[i]; LL res=dinic(); printf("%lld\n",ans-res); return 0; }