最大流裸题,紫书上的图有问题,差点坑到我。。
给出Dinic算法模板,比较高效的最大流算法,复杂度为O(v^2*E),而实际上Dinic算法比这个理论界要好得多。 紫书上的Edmonds-Karp算法的复杂度是O(v*E^2),对于边较多的题目来说显然不够高效。
细节参见代码:
#include<bits/stdc++.h> using namespace std; typedef long long ll; const int INF = 1000000000; const int maxn = 300 + 10; int T,cnt,a,b,s,t,kase = 0,v,c,n; struct Edge { int from, to, cap, flow; }; bool operator < (const Edge& a, const Edge& b) { return a.from < b.from || (a.from == b.from && a.to < b.to); } struct Dinic { int n, m, s, t; vector<Edge> edges; // 边数的两倍 vector<int> G[maxn]; // 邻接表,G[i][j]表示结点i的第j条边在e数组中的序号 bool vis[maxn]; // BFS使用 int d[maxn]; // 从起点到i的距离 int cur[maxn]; // 当前弧指针 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}); m = edges.size(); G[from].push_back(m-2); G[to].push_back(m-1); } bool BFS() { memset(vis, 0, sizeof(vis)); queue<int> Q; Q.push(s); vis[s] = 1; d[s] = 0; 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; d[e.to] = d[x] + 1; Q.push(e.to); } } } return vis[t]; } int DFS(int x, 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(d[x] + 1 == d[e.to] && (f = DFS(e.to, 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) { this->s = s; this->t = t; int flow = 0; while(BFS()) { memset(cur, 0, sizeof(cur)); flow += DFS(s, INF); } return flow; } }g; int main() { while(~scanf("%d",&n)&&n) { scanf("%d%d%d",&s,&t,&c); g.init(n+2); for(int i=1;i<=c;i++) { scanf("%d%d%d",&a,&b,&v); g.AddEdge(a,b,v); g.AddEdge(b,a,v); } printf("Network %d\nThe bandwidth is %d.\n\n",++kase,g.Maxflow(s,t)); } return 0; }