gu 252. Railway Communication （最小费用最大流） 输出方案

参考：http://blog.youkuaiyun.com/wiking__acm/article/details/40185509

#include<cstdio>
#include<map>
#include<queue>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<list>
#include<set>
#include<cmath>
using namespace std;
const int maxn = 1e3 + 5;
const int INF = 1e9;
const double eps = 1e-6;
typedef unsigned long long ULL;
typedef long long LL;
typedef pair<int, int> P;
#define fi first
#define se second
struct Edge {
  int from, to, cap, flow, cost;
};
struct MCMF {
  int n, m, s, t;
  vector<Edge> edges;
  vector<int> G[maxn];
  int inq[maxn];         // 是否在队列中
  int d[maxn];           // Bellman-Ford,单位流量的费用
  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,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;//s-t不连通，失败退出
    flow += a[t];
    cost += d[t] * a[t];
    int u = t;
    while(u != s) {
      edges[p[u]].flow += a[t];
      edges[p[u]^1].flow -= a[t];
      u = edges[p[u]].from;
    }
    return true;
  }

  // 需要保证初始网络中没有负权圈
  int Mincost(int s, int t, int& flow) {
    int cost = 0;
    while(BellmanFord(s, t,flow, cost));
    return cost;
  }
  void print(int tot){
      for(int i=tot;i>=1;--i){
          int e=G[i+tot][0];
          int flow=edges[e].flow;
          if(flow==0){
              int fg=1;
              int pos=i;
              vector<int>path;
              path.push_back(i);
              while(fg){
                  fg=0;
                  for(int i=0;i<G[pos].size();++i){
                      int e=G[pos][i];
                      int flow=edges[e].flow;
                      if(flow!=0&&edges[e].to>tot){
                          fg=1;
                          pos=edges[e].to-tot;
                          path.push_back(pos);
                          break;
                      }
                  }
              }
              printf("%d",path.size());
              for(int i=0;i<path.size();++i){
                  printf(" %d",path[i]);
              }
              puts("");
          }
      }
  }
};

MCMF solver;

int main(){
    int n, m;
    while(scanf("%d%d", &n, &m) != EOF){
        solver.init(2*n+5);
        int source = 0, sink = 2*n+1;
        for(int i = 1;i <= n;i++){
            solver.AddEdge(source, i, 1, 0);
            solver.AddEdge(i+n, sink, 1, 0);
        }
        while(m--){
            int x, y, c;
            scanf("%d%d%d", &x, &y, &c);
            solver.AddEdge(x, y+n, 1, c);
        }
        int flow = 0;
        int cost = solver.Mincost(source, sink, flow);
        printf("%d %d\n", n-flow, cost);
        solver.print(n);
    }
    return 0;
}