2019 Multi-University Training Contest 2 Harmonious Army(最小割)

题意：给你n个点 每个点都有两种选择 成为战士或者法师 现在给你m个关系 对应这两个人的对应关系的权值A,B,C

思路：按照下面的思路建图跑最小割(要注意权值要乘2 可能存在不整除的情况)

#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
const int maxn = 1e3+7;
const int inf = 0x3f3f3f3f;
const double eps = 1e-6;
typedef long long ll;
const ll mod = 1e7+9;
struct Edge {
  ll from, to, cap, flow;
  Edge(int u, int v, int c, int f) : from(u), to(v), cap(c), flow(f) {}
};

struct Dinic {
  int n, m, s, t;
  vector<Edge> edges;
  vector<int> G[maxn];
  int d[maxn], cur[maxn];
  bool vis[maxn];

  void init(int n) {
    for (int i = 0; i < n; i++) G[i].clear();
    edges.clear();
  }

  void AddEdge(int from, int to, int cap) {
    edges.push_back(Edge(from, to, cap, 0));
    edges.push_back(Edge(to, from, 0, 0));
    m = edges.size();
    G[from].push_back(m - 2);
    G[to].push_back(m - 1);
  }

  bool BFS() {
    memset(vis, 0, sizeof(vis));
    queue<int> Q;
    Q.push(s);
    d[s] = 0;
    vis[s] = 1;
    while (!Q.empty()) {
      int x = Q.front();
      Q.pop();
      for (int i = 0; i < G[x].size(); i++) {
        Edge& e = edges[G[x][i]];
        if (!vis[e.to] && e.cap > e.flow) {
          vis[e.to] = 1;
          d[e.to] = d[x] + 1;
          Q.push(e.to);
        }
      }
    }
    return vis[t];
  }

  ll DFS(int x, ll a) {
    if (x == t || a == 0) return a;
    ll flow = 0, f;
    for (int& i = cur[x]; i < G[x].size(); i++) {
      Edge& e = edges[G[x][i]];
      if (d[x] + 1 == d[e.to] && (f = DFS(e.to, min(a, e.cap - e.flow))) > 0) {
        e.flow += f;
        edges[G[x][i] ^ 1].flow -= f;
        flow += f;
        a -= f;
        if (a == 0) break;
      }
    }
    return flow;
  }

  ll Maxflow(int s, int t) {
    this->s = s;
    this->t = t;
    ll flow = 0;
    while (BFS()) {
      memset(cur, 0, sizeof(cur));
      flow += DFS(s, inf);
    }
    return flow;
  }
}dinic;
int main(){
    int n,m;
    while(~scanf("%d%d",&n,&m)){
        dinic.init(n+2);
        ll sum=0;
        for(int i=1;i<=m;i++){
            int u,v;
            ll A,B,C;
            scanf("%d%d%lld%lld%lld",&u,&v,&A,&B,&C);
            sum+=2*(A+B+C);
            ll a,b,c,d,e;
            a=b=(A+B);
            c=d=(B+C);
            e=-2*B+(A+C);
            dinic.AddEdge(0,u,a);
            dinic.AddEdge(0,v,b);
            dinic.AddEdge(u,v,e);
            dinic.AddEdge(v,u,e);
            dinic.AddEdge(u,n+1,c);
            dinic.AddEdge(v,n+1,d);
        }
        printf("%lld\n",(sum-dinic.Maxflow(0,n+1))/2);
    } 
}