UVa 1349 (二分图最小权完美匹配)

题目链接：https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=4095

由于每个点只属于一个有向圈，也就是每个点都只有一个唯一的后继
反之，每个点都有一个唯一的后继，每个点也恰好属于一个圈

于是想到二分图，将每个点 (i) 拆成 (X_i) 和 (Y_i)，有向边 ((u,v)) 对应 (X_u) -> (Y_v), 求出最小权完美匹配即可

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

const int maxn = 505;
const int INF = 1000000007;

int n;

int h[maxn], cnt = 1;
struct E{
	int from, to, cost, cap, next;
}e[maxn * maxn * 100];
void add(int u, int v, int w, int c){
	e[++cnt].to = v;
	e[cnt].from = u;
	e[cnt].cost = w;
	e[cnt].cap = c;
	e[cnt].next = h[u];
	h[u] = cnt;
}

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
  bool BellmanFord(int s, int t, int& flow, int& cost) {
    for(int i = 1 ; i <= n + n + 2 ; ++i) d[i] = INF;
    memset(inq, 0, sizeof(inq));
    memset(a, 0, sizeof(a));
    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 = h[u]; i != -1 ; i = e[i].next) {
        if(e[i].cap && d[e[i].to] > d[u] + e[i].cost) {
          d[e[i].to] = d[u] + e[i].cost;
          p[e[i].to] = i;
          a[e[i].to] = min(a[u], e[i].cap);
          if(!inq[e[i].to]) { Q.push(e[i].to); inq[e[i].to] = 1; }
        }
      }
    }
    if(d[t] == INF) return false;
    
    flow += a[t];
    cost += d[t] * a[t];
    for(int u = t; u != s; u = e[p[u]].from) {
      e[p[u]].cap -= a[t];
      e[p[u]^1].cap += a[t];
    }
    return true;
  }

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


ll read(){ ll s = 0, f = 1; char ch = getchar(); while(ch < '0' || ch > '9'){ if(ch == '-') f = -1; ch = getchar(); } while(ch >= '0' && ch <= '9'){ s = s * 10 + ch - '0'; ch = getchar(); } return s * f; }

int main(){
	while(scanf("%d", &n) == 1 && n){
		memset(h, -1, sizeof(h)); cnt = 1;
		
		int u, v;
		for(int i = 1 ; i <= n ; ++i){
			while(scanf("%d", &u) == 1 && u){
				scanf("%d", &v);
				add(i, u + n, v, 1);
				add(u + n, i, -v, 0);
			}
		}
		
		int s = n + n + 1, t = n + n + 2;
		for(int i = 1 ; i <= n ; ++i){ // 源点 -> X 
			add(s, i, 0, 1);
			add(i, s, 0, 0);
		}
		for(int i = 1 ; i <= n ; ++i){ // Y -> 汇点 
			add(i + n, t, 0, 1);
			add(t, i + n, 0, 0);
		}
		
		int flow = 0, cost = 0;
		flow = MincostFlow(s, t, cost);
		
		if(flow == n){
			printf("%d
", cost);
		} else{
			printf("N
");
		}
	}
	return 0;
}