Created
July 25, 2012 18:11
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simple implementation of Dinic's algorithm for solving maxflow problem
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#define MAX 500 | |
class Dinic { | |
int n, m, head[MAXN], level[MAXN], s, t, work[MAXN]; | |
struct edge { | |
int v, c, f, nxt; | |
edge() {} | |
edge(int v, int c, int f, int nxt): v(v), c(c), f(f), nxt(nxt) {} | |
} e[MAXM]; | |
bool _bfs() { | |
static int q[MAXN]; | |
memset(level, -1, sizeof(int) * n); | |
int le = 0, ri = 0; | |
q[ri++] = s; | |
level[s] = 0; | |
while(le < ri) { | |
for(int k = q[le++], i = head[k]; i != -1; i = e[i].nxt) { | |
if(e[i].f < e[i].c && level[e[i].v] == -1) { | |
level[e[i].v] = level[k] + 1; | |
q[ri++] = e[i].v; | |
} | |
} | |
} | |
return (level[t] != -1); | |
} | |
int _dfs(int u, int f) { | |
if(u == t) | |
return f; | |
for(int& i = work[u]; i != -1; i = e[i].nxt) { | |
if(e[i].f < e[i].c && level[u] + 1 == level[e[i].v]) { | |
int minf = _dfs(e[i].v, min(f, e[i].c - e[i].f)); | |
if(minf > 0) { | |
e[i].f += minf; | |
e[i ^ 1].f -= minf; | |
return minf; | |
} | |
} | |
} | |
return 0; | |
} | |
public: | |
void init(int nn, int src, int dst) { | |
n = nn; | |
s = src; | |
t = dst; | |
m = 0; | |
memset(head, -1, sizeof(int) * n); | |
} | |
void addEdge(int u, int v, int c, int rc) { | |
assert(u < n); | |
assert(v < n); | |
e[m] = edge(v, c, 0, head[u]); | |
head[u] = m++; | |
e[m] = edge(u, rc, 0, head[v]); | |
head[v] = m++; | |
assert(m < MAXM); | |
} | |
int maxFlow() { | |
int ret = 0; | |
while(_bfs()) { | |
memcpy(work, head, sizeof(int) * n); | |
while(true) { | |
int delta = _dfs(s, INF); | |
if(delta == 0) | |
break; | |
ret += delta; | |
} | |
} | |
return ret; | |
} | |
}; |
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