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;
}