UVa 1349 Optimal Bus Route Design (最佳完美匹配)

题意：给定一个有向图，让你找出若干个图，使得每个点恰好属于一个圈，并且总的权和最小。

析：每个点都有唯一的一个圈，也就是说每一点都有唯一的后继，那么我们就可以转换成求一个图的最小权的最佳完全匹配，可以用最小费用流来求，

先把每个结点拆成两个点，假设是x，y，然后建立一个源点，向每个点的x连一条容量为1的边，建立一个汇点，每个点的y向汇点连一条容量为1的边，

每条边u,v，也连接一条容量为1，费用为权值的边，最小求一个最小费用流即可。

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>
#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 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 double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-5;
const int maxn = 2000 + 10;
const int mod = 1e6;
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;
  LL cost;
};

struct MCMF{
  int n, m;
  vector<Edge> edges;
  vector<int> G[maxn];
  int inq[maxn];
  LL d[maxn];
  int p[maxn];
  int a[maxn];

  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, LL 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, LL &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;
    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;
  }

  int minCost(int s, int t, LL &cost){
    int flow = 0;  cost = 0;
    while(BellmanFord(s, t, flow, cost));
    return flow;
  }
};
MCMF mcmf;

int main(){
  while(scanf("%d", &n) == 1 && n){
    mcmf.init(n+n+2);
    int v;  LL val;
    int s = 0, t = n+n+1;
    for(int i = 1; i <= n; ++i){
     // mcmf.addEdge(i, i+n, 1, 0);
      mcmf.addEdge(s, i, 1, 0);
      mcmf.addEdge(i+n, t, 1, 0);
      while(scanf("%d", &v) == 1 && v){
        scanf("%lld", &val);
        mcmf.addEdge(i, v+n, 1, val);
      }
    }

    LL ans;
    if(mcmf.minCost(s, t, ans) == n)  printf("%lld\n", ans);
    else  printf("N\n");
  }
  return 0;
}