Dinic - NOI2006 maximum profit

Modeling:

最大权闭合子图。
对于每一个项目，将它与两个中转站连INF边，表示他们之间的关系。
从s与每个项目连边，流量为项目的获利，表示不选它会损失的代价。
每个中转站与t连边，表示选它会花费的代价。

2007 can be seen minimal cut paper

No current arc plus optimization, has been TLE. . .
Dinic attached template (the current arc optimization)

#include<cstdio>
#include<cstring>
#include<iostream>
#define For(aa,bb,cc) for(int aa=(bb);aa<=(int)(cc);++aa)
using namespace std;
const int maxn=1e6+10,inf=2147483647;
int n,m,s,t;
int be[maxn],ne[maxn],to[maxn],flow[maxn],e;
int dis[maxn],cur[maxn];

inline void add_edge(int x,int y,int z){
    to[++e]=y,ne[e]=be[x],be[x]=e,flow[e]=z;
    to[++e]=x,ne[e]=be[y],be[y]=e,flow[e]=0;
}

int bfs(){
    For(i,1,n+m+2) dis[i]=-1;
    int q[maxn],l=0,r=0;
    q[++r]=s,dis[s]=0;
    while(l<r){
        int k=q[++l];
        for(int i=be[k];i;i=ne[i]){
            int u=to[i];
            if(flow[i] && dis[u]==-1){
                dis[u]=dis[k]+1;
                q[++r]=u;
            }
        }
    }
    return dis[t]!=-1;
}

int dfs(int node,int f){
    if(t==node) return f;
    int res;
    for(int &i=cur[node];i;i=ne[i]){
        int u=to[i];
        if(flow[i] && dis[u]==dis[node]+1){
            if((res=dfs(u,min(f,flow[i])))){
                flow[i]-=res;
                flow[i^1]+=res;
                return res;
            }
        }
    }
    return 0;
}

int Dinic(){
    int ans=0,tmp;
    while(bfs()){
        memcpy(cur,be,sizeof be);
        while((tmp=dfs(s,inf))) ans+=tmp;
    }
    return ans;
}

int main(){
#ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    freopen("out.txt","w",stdout);
#endif
    int x,y,z,res=0;
    e=1;
    scanf("%d%d",&n,&m);
    s=n+m+1,t=n+m+2;
    For(i,1,n) scanf("%d",&x),add_edge(i,t,x);
    For(i,1,m){
        scanf("%d%d%d",&x,&y,&z);
        add_edge(i+n,x,inf);
        add_edge(i+n,y,inf);
        add_edge(s,i+n,z);
        res+=z;
    }
    printf("%d\n",res-Dinic());
    return 0;
}