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LibreOJ #6000. 「网络流 24 题」搭配飞行员
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发布时间:2023-01-31

本文共 3788 字,大约阅读时间需要 12 分钟。

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/*    LibreOJ #6000. 「网络流 24 题」搭配飞行员        二分图最大匹配    Dinic最大流 + 当前弧优化     */#include 
#include
#include
#define Max 10000#define INF 1e5int read (int &now){ now = 0; register char word = getchar (); while (word < '0' || word > '9') word = getchar (); while (word >= '0' && word <= '9') { now = now * 10 + word - '0'; word = getchar (); } if (now >= 0) return 1;}inline int min (int a, int b){ return a < b ? a : b;}class Net_Flow_Type{ private : int __to[Max << 2], __next[Max << 2]; int __flow[Max << 2]; int edge_list[Max]; int Edge_Count; int deep[Max], __tech_[Max]; int T; int Answer; public : Net_Flow_Type () { Edge_Count = 1; } inline void Insert_edge (int from, int to) { Edge_Count ++; __to[Edge_Count] = to; __next[Edge_Count] = edge_list[from]; edge_list[from] = Edge_Count; Edge_Count ++; __to[Edge_Count] = from; __next[Edge_Count] = edge_list[to]; edge_list[to] = Edge_Count; __flow[Edge_Count - 1] = 1; __flow[Edge_Count] = 0; } bool Bfs (int Start, int End) { std :: queue
Queue; Queue.push (Start); memset (deep, -1, sizeof deep); int now; for (deep[Start] = 0; !Queue.empty (); Queue.pop ()) { now = Queue.front (); for (int i = edge_list[now]; i; i = __next[i]) if (__flow[i] && deep[__to[i]] == -1) { deep[__to[i]] = deep[now] + 1; if (__to[i] == End) return true; Queue.push (__to[i]); } } return deep[End] != -1; } int Flowing (int now, int flow) { if (now == T || flow <= 0) return flow; int res = 0, pos = 0; for (int i = __tech_[now]; i; i = __next[i]) { if (deep[__to[i]] != deep[now] + 1 || __flow[i] <= 0) continue; res = Flowing (__to[i], min (flow, __flow[i])); if (res > 0) { flow -= res; pos += res; __flow[i] -= res; __flow[i ^ 1] += res; if (__flow[i]) __tech_[now] = i; if (flow == 0) return pos; } } return pos; } int Dinic (int Start, int End) { for (T = End; Bfs (Start, End); ) { memcpy (__tech_, edge_list, sizeof edge_list); Answer += Flowing (Start, INF); } return Answer; } };int N, M;Net_Flow_Type Make;int main (int argc, char *argv[]){ read (N); read (M); int S = N + 1, T = N + 2; for (int i = 1; i <= M; i ++) Make.Insert_edge (S, i); for (int i = M + 1; i <= N; i ++) Make.Insert_edge (i, T); for (int x, y; scanf ("%d %d", &x, &y) == 2; Make.Insert_edge (x, y)); printf ("%d", Make.Dinic (S, T)); return 0;}

 

转载于:https://www.cnblogs.com/ZlycerQan/p/7076874.html

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