二分图网络流做法
(1)最大基数匹配。源点到每一个X节点连一条容量为1的弧, 每一个Y节点连一条容量为1的弧, 然后每条有向
边连一条弧, 容量为1, 然后跑一遍最大流即可, 最大流即是最大匹配对数
(2)最小(大)权完美匹配(每个点都被匹配到)。和最大基数匹配类似, 只是有向边的权值就是费用, 其余弧费用为0.
跑一遍最小费用流。最后要判断从s出发的弧是否满载, 不是则不能完美匹配。如果求最大权那么费用设为负的就ok。
这道题目每一个点恰好在一个圈内, 也就是说每一个点只有唯一的后继。反过来, 如果每一个点只有唯一的后继
因为要二分图, 所以拆点, 每个点拆成xi和yi, 然后a与b连接的时候xa连yb, 这样就变成了二分图最小权完美匹配。
#include<cstdio> #include<vector> #include<queue> #include<algorithm> #include<cstring> #define REP(i, a, b) for(int i = (a); i < (b); i++) using namespace std; typedef long long ll; const int MAXN = 212; struct Edge { int from, to, cap, flow, cost; Edge(int from, int to, int cap, int flow, int cost) : from(from), to(to), cap(cap), flow(flow), cost(cost) {}; }; vector<Edge> edges; vector<int> g[MAXN]; int p[MAXN], a[MAXN], d[MAXN], vis[MAXN], n, m, s, t; 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)); g[from].push_back(edges.size() - 2); g[to].push_back(edges.size() - 1); } bool spfa(int& flow, ll& cost) { REP(i, 0, t + 1) d[i] = (i == s ? 0 : 1e9); memset(vis, 0, sizeof(vis)); a[s] = 1e9; vis[s] = 1; p[s] = 0; queue<int> q; q.push(s); while(!q.empty()) { int u = q.front(); q.pop(); vis[u] = 0; REP(i, 0, g[u].size()) { 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(!vis[e.to]) { vis[e.to] = 1; q.push(e.to); } } } } if(d[t] == 1e9) return false; 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; } int mincost(ll& cost) { int flow = 0; cost = 0; while(spfa(flow, cost)); return flow; } int main() { while(~scanf("%d", &n) && n) { s = 0; t = 2 * n + 1; REP(i, 0, t + 1) g[i].clear(); edges.clear(); for(int i = 1; i <= n; i++) { AddEdge(s, i, 1, 0); AddEdge(n + i, t, 1, 0); } for(int i = 1; i <= n; i++) { while(1) { int j, d; scanf("%d", &j); if(j == 0) break; scanf("%d", &d); AddEdge(i, n + j, 1, d); } } ll ans, flow; flow = mincost(ans); if(flow != n) puts("N"); else printf("%lld\n", ans); } return 0; }