网络流

网络流专题：点击打开链接

一、最大流之增广路算法

这个博客讲解放入挺详细的：点击打开链接

前向弧：离开点u的有向弧

后向弧：进入点u的有向弧

【恩~有一点我觉得有必要记录一下】

建立这些后向弧的必要性：

如果不建立后向边就容易出现上面这种情况，结果会偏小，路径是：1->2->5、1->2->3->5、1->3->5。

建立后向边之后，路径会增加一条：1->2->5、1->2->3->5、1->3->2->4->5、1->3->5。

2->3流量是1，3->2流量是1，相当于就没有流过。

模板题：【hdu 1532】

1、EdmondsKarp

struct Edge {int from, to, cap, flow;Edge(int u, int v, int c, int f):from(u),to(v),cap(c),flow(f){}}; struct EdmondsKarp {int n, m;vector<Edge> edges;vector<int> G[maxn];int a[maxn];int p[maxn];void init(int n) {for(int i=0; i<n; i++) G[i].clear();edges.clear();}void AddEdge(int from, int to, int cap) {edges.push_back(Edge(from, to, cap, 0));edges.push_back(Edge(to, from, 0, 0));//反向弧m  = edges.size();G[from].push_back(m-2);G[to].push_back(m-1);}int Maxflow(int s, int t) {int flow = 0;for(;;){memset(a, 0, sizeof(a));queue<int>Q;Q.push(s);a[s] = INF;while(!Q.empty()) {int x = Q.front(); Q.pop();int len = (int)G[x].size();for(int i=0; i<len; i++) {Edge& e = edges[G[x][i]];if(!a[e.to] && e.cap>e.flow) {p[e.to] = G[x][i];a[e.to] = min(a[x], e.cap-e.flow);Q.push(e.to);}}if(a[t]) break;}if(!a[t]) break;for(int u = t; u!=s; u=edges[p[u]].from) {edges[p[u]].flow += a[t];edges[p[u]^1].flow -= a[t];}flow += a[t];}return flow;}};

2、Dinic递归：

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <string>#include <vector>#include <queue>using namespace std;const int maxn=1e3+10;const int INF = 0x3f3f3f3f;struct Edge {int from, to, cap, flow;Edge(int u, int v, int c, int f):from(u),to(v),cap(c),flow(f){}}; vector<Edge> edges;vector<int>G[maxn];bool vis[maxn];int dis[maxn];int cur[maxn];void init(int n) {for(int i=0; i<n; i++) G[i].clear();edges.clear();}void AddEdge(int from, int to, int cap) {edges.push_back(Edge(from, to, cap, 0));edges.push_back(Edge(to, from, 0, 0));//反向弧int m  = edges.size();G[from].push_back(m-2);G[to].push_back(m-1);}bool BFS(int s, int t) {memset(vis, 0, sizeof(vis));int n, m;queue<int> Q;Q.push(s);dis[s] = 0;vis[s] = 1;while(!Q.empty()) {int x = Q.front(); Q.pop();for(int i = 0; i < G[x].size(); i++) {Edge& e = edges[G[x][i]];if(!vis[e.to] && e.cap > e.flow) {vis[e.to] = 1;dis[e.to] = dis[x] + 1;Q.push(e.to);}}}return vis[t];}int DFS(int x, int t, int a){if(x == t || a == 0) return a;int flow = 0, f;for(int& i = cur[x]; i < G[x].size(); i++) {Edge& e = edges[G[x][i]];if(dis[x] + 1 == dis[e.to] && (f = DFS(e.to, t, min(a, e.cap-e.flow))) > 0) {e.flow += f;edges[G[x][i]^1].flow -= f;flow += f;a -= f;if(a == 0) break;}}return flow;}int Maxflow(int s, int t){int flow = 0;while(BFS(s, t)) {memset(cur, 0, sizeof(cur));flow += DFS(s, t, INF);}return flow;}int main(){int m, n;while(~scanf("%d%d", &m, &n)) {init(n);for(int i=0; i<m; i++){int u, v, c;scanf("%d%d%d", &u, &v, &c);AddEdge(u, v, c);}printf("%d\n", Maxflow(1, n));}return 0;}

3、Dinic非递归：

#include <cstdio>  #include <cstring>  #include <algorithm>  #include <iostream>  using namespace std;  //点标 [0,n]  const int N = 200010;  const int M = 500010;  const int INF = ~0u >> 2;  template<class T>  struct Max_Flow {      int n;      int Q[N], sign;      int head[N], level[N], cur[N], pre[N];      int nxt[M], pnt[M], E;      T cap[M];      void Init(int n) {          this->n = n+1;          E = 0;          std::fill(head, head + this->n, -1);      }      //有向rw 就= 0       void add(int from, int to, T c, T rw) {          pnt[E] = to;          cap[E] = c;          nxt[E] = head[from];          head[from] = E++;            pnt[E] = from;          cap[E] = rw;          nxt[E] = head[to];          head[to] = E++;      }      bool Bfs(int s, int t) {          sign = t;          std::fill(level, level + n, -1);          int *front = Q, *tail = Q;          *tail++ = t; level[t] = 0;          while(front < tail && level[s] == -1) {              int u = *front++;              for(int e = head[u]; e != -1; e = nxt[e]) {                  if(cap[e ^ 1] > 0 && level[pnt[e]] < 0) {                      level[pnt[e]] = level[u] + 1;                      *tail ++ = pnt[e];                  }              }          }          return level[s] != -1;      }      void Push(int t, T &flow) {          T mi = INF;          int p = pre[t];          for(int p = pre[t]; p != -1; p = pre[pnt[p ^ 1]]) {              mi = std::min(mi, cap[p]);          }          for(int p = pre[t]; p != -1; p = pre[pnt[p ^ 1]]) {              cap[p] -= mi;              if(!cap[p]) {                  sign = pnt[p ^ 1];              }              cap[p ^ 1] += mi;          }          flow += mi;      }      void Dfs(int u, int t, T &flow) {          if(u == t) {              Push(t, flow);              return ;          }          for(int &e = cur[u]; e != -1; e = nxt[e]) {              if(cap[e] > 0 && level[u] - 1 == level[pnt[e]]) {                  pre[pnt[e]] = e;                  Dfs(pnt[e], t, flow);                  if(level[sign] > level[u]) {                      return ;                  }                  sign = t;              }          }      }      T Dinic(int s, int t) {          pre[s] = -1;          T flow = 0;          while(Bfs(s, t)) {              std::copy(head, head + n, cur);              Dfs(s, t, flow);          }          return flow;      }  };  Max_Flow <int>F;  int main(){      int t, n, m;scanf("%d", &t);      while(t--) {      scanf("%d%d", &n, &m);        F.Init(n);          int s = 1, t = n;        for(int i=0; i<m; i++){              int u, v, c;              scanf("%d%d%d", &u, &v, &c);              F.add(u, v, c, c);          }          printf("%d\n", F.Dinic(s, t));      }      return 0;  }  

【HDU 4280】 Island Transport 用Dinic非递归做能对，递归超时

易错例题：

A、【POJ 3281 Dining】

有n头牛，f中食物，d种饮料，每头牛都只吃固定的几种食物跟饮料，问最多有几头牛能吃到一种食物，一种饮料

注意点就是建图的时候n个点要拆成2*n个点，免得一头牛吃了多食物跟饮料。

二、最小割最大流

三、最小费用最大流

和Edmonds-Karp类似，但每次用Bellman-Ford算法而非BFS找增广路。

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <string>#include <vector>#include <queue>#include <map>using namespace std;const int maxn=1e3+10;const int INF = 0x3f3f3f3f;struct Edge {int from, to, cap, flow, cost;Edge(int u, int v, int c, int f, int w):from(u), to(v), cap(c), flow(f), cost(w){}};vector<Edge> edges;vector<int> G[maxn];int inq[maxn];int d[maxn];int p[maxn];int a[maxn];void init(){for(int i=0; i<maxn; i++) G[i].clear();edges.clear();}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));int m = edges.size();G[from].push_back(m-2);G[to].push_back(m-1);}bool BellmanFord(int s, int t, int& flow, long long& cost) {memset(inq, 0, sizeof(inq));memset(d, INF, sizeof(d));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 += (long long)d[t] * (long long)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;}//需要保证初始网络中没有负权圈int MincostMaxflow(int s, int t, long long& cost) {int flow = 0;cost = 0;while(BellmanFord(s, t, flow, cost));return flow;}int main() {int n, m;while(~scanf("%d%d", &n, &m)){init();for(int i=0; i<m; i++){int u, v, f, c;scanf("%d%d%d%d", &u, &v, &f, &c);AddEdge(u, v, f, c);}long long cost=0;int flow = MincostMaxflow(1, n, cost);printf("%d %lld\n", flow, cost);}return 0;}