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<bits/stdc++.h> 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<int,int> 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<int> 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; }