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UVaLive 4597 Inspection (网络流,最小流)

题意:给出一张有向图,每次你可以从图中的任意一点出发,经过若干条边后停止,然后问你最少走几次可以将图中的每条边都走过至少一次,并且要输出方案,这个转化为网络流的话,就相当于 求一个最小流,并且存在下界,即每条边至少走一次。

析:转载:http://blog.csdn.net/sdj222555/article/details/40380423

这上面说的很清楚了,就是先求一遍是正向的,然后再求逆向的,这个逆向就是为了求可以减少的边。

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>
#include <numeric>
#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<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-3;
const int maxn = 100 + 40;
const int maxm = (100000 << 8) + 10;
const int mod = 1000000007;
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];
  int d[maxn];
  bool vis[maxn];
  int cur[maxn];

  void init(int n){
    this-> n = n;
    FOR(i, 0, n)  G[i].cl;
    edges.cl;
  }

  void addEdge(int from, int to, int cap){
    edges.pb((Edge){from, to, cap, 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 in[maxn];
vector<P>  edges[maxn];
int main(){
  while(scanf("%d", &n) == 1){
    ms(in, 0);
    int s = 0, t = n + 1;
    dinic.init(t + 4);
    int ans = 0;
    for(int i = 1; i <= n; ++i){
      int x;  scanf("%d", &x);
      while(x--){
        int y;  scanf("%d", &y);
        ++in[y];  --in[i];
        dinic.addEdge(i, y, INF);
      }
      edges[i].cl;
    }
    int sum = 0;
    for(int i = 1; i <= n; ++i)
      if(in[i] > 0)  dinic.addEdge(s, i, in[i]), ans += in[i];
      else  dinic.addEdge(i, t, -in[i]);
    printf("%d\n", ans -= dinic.maxflow(s, t));
    for(int i = 1; i <= n; ++i){
      for(int j = 0; j < dinic.G[i].sz; ++j){
        Edge &e = dinic.edges[dinic.G[i][j]];
        if(e.from != i || e.to == t || e.cap == 0)  continue;
        edges[i].pb(P(e.to, e.flow + 1));
        in[e.to] += e.flow;
      }
    }
    for(int i = 1; i <= n; ++i)  while(in[i] < 0){
      ++in[i];
      vector<int> ans;  ans.push_back(i);
      int u = i;
      while(1){
        bool ok = true;
        for(int j = 0; j < edges[u].sz; ++j){
          if(edges[u][j].se == 0)  continue;
          ok = false;
          --edges[u][j].se;
          u = edges[u][j].fi;
          ans.push_back(u);
          break;
        }
        if(ok)   break;
      }
      for(auto &it : ans)  printf("%d%c", it, " \n"[it == ans.back()]);
    }
  }
  return 0;
}

  

posted on 2017-11-03 22:34 dwtfukgv 阅读(...) 评论(...) 编辑 收藏