将第i个用户和他需要的基站连边，转化成求二分图的最大权闭合子图。
答案=正权点之和-最小割。
# include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # include <vector> # include <queue> # include <stack> # include <map> # include <set> # include <cmath> # include <algorithm> using namespace std; # define lowbit(x) ((x)&(-x)) # define pi 3.1415926535 # define eps 1e-9 # define MOD 1000000009 # define INF 1000000000 # 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 res=0, flag=0; char ch; if((ch=getchar())=='-') flag=1; else if(ch>='0'&&ch<='9') res=ch-'0'; while((ch=getchar())>='0'&&ch<='9') res=res*10+(ch-'0'); return flag?-res:res; } void Out(int a) { if(a<0) {putchar('-'); a=-a;} if(a>=10) Out(a/10); putchar(a%10+'0'); } const int N=60005; //Code begin... struct Edge{int p, next, w;}edge[350005]; int head[N], cnt=2, s, t, vis[N]; queue<int>Q; void add_edge(int u, int v, int 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 (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; } int dfs(int x, int low){ int i, a, temp=low; if (x==t) return low; for (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; } int main () { int n, m, u, v, tmp; while (~scanf("%d%d",&n,&m)) { s=0, t=n+m+1; int res=0, sum=0; mem(head,0); cnt=2; FOR(i,1,n) scanf("%d",&tmp), add_edge(m+i,t,tmp); FOR(i,1,m) { scanf("%d%d%d",&u,&v,&tmp); res+=tmp; add_edge(s,i,tmp); add_edge(i,m+u,INF); add_edge(i,m+v,INF); } while (bfs()) while (tmp=dfs(s,INF)) sum+=tmp; printf("%d ",res-sum); } return 0; }