uva1658

Posted 2021-01-24 033000-

因为直接利用费用流可能会导致经过相同的点，所以利用拆点法

除了首尾结点之外其他的每一个结点都拆成两个结点，并且两个结点之间连一条容量为1，费用为0的边，这样做的理由也很简单

一旦一个节点用过之后，下次再经过这个节点之后再也没有办法扩展到其他节点了

#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iomanip>
#include<assert.h>
#include<ctime>
#include<vector>
#include<list>
#include<map>
#include<set>
#include<sstream>
#include<stack>
#include<queue>
#include<string>
#include<bitset>
#include<algorithm>
using namespace std;
#define me(s)  memset(s,0,sizeof(s))
#define pf printf
#define sf scanf
#define Di(x) int x;scanf("%d",&x)
#define in(x) inp(x)
#define in2(x,y) inp(x),inp(y)
#define in3(x,y,z) inp(x),inp(y),inp(z)
#define ins(x) scanf("%s",x)
#define ind(x) scanf("%lf",&x)
#define IO ios_base::sync_with_stdio(0);cin.tie(0)
#define READ freopen("C:/Users/ASUS/Desktop/in.txt","r",stdin)
#define WRITE freopen("C:/Users/ASUS/Desktop/out.txt","w",stdout)
template<class T> void inp(T &x) {//读入优化
    char c = getchar(); x = 0;
    for (; (c < 48 || c>57); c = getchar());
    for (; c > 47 && c < 58; c = getchar()) { x = (x << 1) + (x << 3) + c - 48; }
}
typedef pair <int, int> pii;
typedef long long ll;
typedef unsigned long long ull;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-15;
const int maxn=2000+5;
struct Edge{
    int from,to,cap,flow,cost;
    Edge(int u,int v,int c,int f,int w):from(u),to(v),cap(c),flow(f),cost(w){}
};
struct MCMF {
  int n, m;
  vector<Edge> edges;
  vector<int> G[maxn];
  int inq[maxn];         // 是否在队列中
  int d[maxn];           // Bellman-Ford
  int p[maxn];           // 上一条弧
  int a[maxn];           // 可改进量

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

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

  bool BellmanFord(int s, int t, int flow_limit, int& flow, int& cost) {
    for(int i = 0; i < n; i++) d[i] = inf;
    memset(inq, 0, sizeof(inq));
    d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = inf;

    queue<int> Q;
    Q.push(s);
    while(!Q.empty()) {
      int u = Q.front(); Q.pop();
      inq[u] = 0;
      for(int i = 0; i < G[u].size(); i++) {
        Edge& e = edges[G[u][i]];
        if(e.cap > e.flow && d[e.to] > d[u] + e.cost) {
          d[e.to] = d[u] + e.cost;
          p[e.to] = G[u][i];
          a[e.to] = min(a[u], e.cap - e.flow);
          if(!inq[e.to]) { Q.push(e.to); inq[e.to] = 1; }
        }
      }
    }
    if(d[t] == inf) return false;
    if(flow + a[t] > flow_limit) a[t] = flow_limit - flow;
    flow += a[t];
    cost += d[t] * a[t];
    for(int u = t; u != s; u = edges[p[u]].from) {
      edges[p[u]].flow += a[t];
      edges[p[u]^1].flow -= a[t];
    }
    return true;
  }

  // 需要保证初始网络中没有负权圈
  int MincostMaxFlow(int s, int t, int flow_limit, int& cost) {
    int flow = 0; cost = 0;
    while(flow < flow_limit && BellmanFord(s, t, flow_limit, flow, cost));
    return flow;
  }

};
MCMF g;
int n,m;
int main(){
    //READ;
    //WRITE;
    //clock_t startTime,endTime;
    //startTime = clock();//计时开始
    while(sf("%d%d",&n,&m)==2){
        g.init(2*n-2);
        for(int i=1;i<n-1;i++)
        g.AddEdge(i,i+n-1,1,0);
    
    int u,v,c;
    while(m--){
        in3(u,v,c);
        if(u!=1&&u!=n) u+=n-2;
        else u--;
        v--;
        g.AddEdge(u,v,1,c);
    }
    int ans=0;
    g.MincostMaxFlow(0,n-1,2,ans);
    cout<<ans<<endl;
    }
    //endTime = clock();//计时结束
    //cout << "The run time is: " <<(double)(endTime - startTime) / CLOCKS_PER_SEC << "s" << endl;
    //system("pause");
}