题意：现在有一个图，选择一条边，会把边的2个顶点也选起来，最后会的到一个边的集合 和一个点的集合 ， 求边的集合 - 点的集合最大是多少。

题解：裸的最大权闭合子图。

代码：

#include
      
       using namespace std;#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);#define LL long long#define ULL unsigned LL#define fi first#define se second#define pb push_back#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define lch(x) tr[x].son[0]#define rch(x) tr[x].son[1]#define max3(a,b,c) max(a,max(b,c))#define min3(a,b,c) min(a,min(b,c))typedef pair
       
         pll;const int inf = 0x3f3f3f3f;const LL INF = 0x3f3f3f3f3f3f3f3f;const LL mod =  (int)1e9+7;const int N = 2e3 + 100, M = 1e5;int head[N], deep[N], cur[N];int w[M], to[M], nx[M];int tot;void add(int u, int v, int val){    w[tot]  = val; to[tot] = v;    nx[tot] = head[u]; head[u] = tot++;    w[tot] = 0; to[tot] = u;    nx[tot] = head[v]; head[v] = tot++;}int bfs(int s, int t){    queue
        
          q;    memset(deep, 0, sizeof(deep));    q.push(s);    deep[s] = 1;    while(!q.empty()){        int u = q.front();        q.pop();        for(int i = head[u]; ~i; i = nx[i]){            if(w[i] > 0 && deep[to[i]] == 0){                deep[to[i]] = deep[u] + 1;                q.push(to[i]);            }        }    }    return deep[t] > 0;}int Dfs(int u, int t, int flow){    if(u == t) return flow;    for(int &i = cur[u]; ~i; i = nx[i]){        if(deep[u]+1 == deep[to[i]] && w[i] > 0){            int di = Dfs(to[i], t, min(w[i], flow));            if(di > 0){                w[i] -= di, w[i^1] += di;                return di;            }        }    }    return 0;}LL Dinic(int s, int t){    LL ans = 0, tmp;    while(bfs(s, t)){        for(int i = 0; i <= t; i++) cur[i] = head[i];        while(tmp = Dfs(s, t, inf)) ans += tmp;    }    return ans;}void init(){    memset(head, -1, sizeof(head));    tot = 0;}int main(){    int n, m, s, t;    scanf("%d%d", &n, &m);    init();    s = 0; t = m+n+1;    for(int i = 1, v; i <= n; ++i){        scanf("%d", &v);        add(i,t,v);    }    LL ans = 0;    for(int i = 1, v, u, w; i <= m; ++i){        scanf("%d%d%d", &u, &v, &w);        add(n+i, u, inf);        add(n+i, v, inf);        add(s, n+i, w);        ans += w;    }    ans -= Dinic(s,t);    printf("%lld\n", ans);    return 0;}