HDU 3897 Base Station (网络流，最大闭合子图)

题意：给定n个带权点m条无向带权边，选一个子图，则这个子图的权值为 边权和-点权和，求一个最大的权值。

析：把每条边都看成是一个新点，然后建图，就是一个裸的最大闭合子图。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <sstream>

#include <list>

#include <assert.h>

#include <bitset>

#define debug() puts("++++");

#define gcd(a, b) __gcd(a, b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define fi first

#define se second

#define pb push_back

#define sqr(x) ((x)*(x))

#define ms(a,b) memset(a, b, sizeof a)

#define sz size()

#define pu push_up

#define pd push_down

#define cl clear()

#define all 1,n,1

#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<LL, int> P;

const int INF = 0x3f3f3f3f;

const LL LNF = 1e17;

const double inf = 1e20;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 55000 + 50;

const int maxm = 1e6 + 5;

const int mod = 10007;

const int dr[] = {-1, 0, 1, 0};

const int dc[] = {0, -1, 0, 1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline bool is_in(int r, int c) {

  return r >= 0 && r < n && c >= 0 && c < m;

}

struct Edge{

  int from, to, cap, flow;

};

struct Dinic{

  int n, m, s, t;

  vector<Edge> edges;

  vector<int> G[maxn];

  bool vis[maxn];

  int d[maxn];

  int cur[maxn];

  void init(int n){

    this->n = n;

    for(int i = 0; i < n; ++i)  G[i].cl;

    edges.cl;

  }

  void addEdge(int from, int to, int c){

    edges.pb((Edge){from, to, c, 0});

    edges.pb((Edge){to, from, 0, 0});

    m = edges.sz;

    G[from].pb(m - 2);

    G[to].pb(m - 1);

  }

  bool bfs(){

    ms(vis, 0);  vis[s] = 1;  d[s] = 0;

    queue<int> q;

    q.push(s);

    while(!q.empty()){

      int u = q.front();  q.pop();

      for(int i = 0; i < G[u].sz; ++i){

        Edge &e = edges[G[u][i]];

        if(!vis[e.to] && e.cap > e.flow){

          d[e.to] = d[u] + 1;

          vis[e.to] = 1;

          q.push(e.to);

        }

      }

    }

    return vis[t];

  }

  int dfs(int u, int a){

    if(u == t || a == 0)  return a;

    int flow = 0, f;

    for(int &i = cur[u]; i < G[u].sz; ++i){

      Edge &e = edges[G[u][i]];

      if(d[e.to] == d[u] + 1 && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0){

        e.flow += f;

        edges[G[u][i]^1].flow -= f;

        flow += f;

        a -= f;

        if(a == 0)  break;

      }

    }

    return flow;

  }

  int maxflow(int s, int t){

    this->s = s;   this->t = t;

    int flow = 0;

    while(bfs()){ ms(cur, 0);  flow += dfs(s, INF); }

    return flow;

  }

};

Dinic dinic;

int main(){

  while(scanf("%d %d", &n, &m) == 2){

    int s = 0, t = n + m + 1;

    dinic.init(t + 5);

    for(int i = 1; i <= n; ++i){

      int c;  scanf("%d", &c);

      dinic.addEdge(i, t, c);

    }

    int sum = 0;

    for(int i = 1; i <= m; ++i){

      int u, v, c;

      scanf("%d %d %d", &u, &v, &c);

      dinic.addEdge(n + i, u, INF);

      dinic.addEdge(n + i, v, INF);

      dinic.addEdge(s, n + i, c);

      sum += c;

    }

    printf("%d\n", sum - dinic.maxflow(s, t));

  }

  return 0;

}