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使用ISAP算法进行二分图匹配(附UOJ - 78 二分图最大匹配代码)
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Bigraph-Matching(Improved-Shortest-Augmenting-Path).cpp
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#include <cstdio> | ||
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#define INF 1000000000 | ||
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using namespace std; | ||
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struct vertex | ||
{ | ||
int first, dis; | ||
}V[1010]; | ||
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struct edge | ||
{ | ||
int endp, next, flow; | ||
}E[502010]; | ||
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int nl, nr, iec, ec = 2, src = 1005, sink = 1006, gap[1010]; | ||
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inline int min(int x, int y) | ||
{ | ||
return x < y ? x : y; | ||
} | ||
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void init() | ||
{ | ||
gap[0] = nl + nr + 2; | ||
} | ||
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void add_edge(int u, int v, int f) | ||
{ | ||
E[ec].next = V[u].first; | ||
V[u].first = ec; | ||
E[ec].endp = v; | ||
E[ec].flow = f; | ||
ec++; | ||
} | ||
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int isap(int u, int curf) | ||
{ | ||
if (u == sink) | ||
{ | ||
return curf; | ||
} | ||
int totalf = 0, mindis = nl + nr + 2; | ||
for (int cur = V[u].first; cur != 0 && totalf < curf; cur = E[cur].next) | ||
{ | ||
if (E[cur].flow > 0) | ||
{ | ||
if (V[u].dis == V[E[cur].endp].dis + 1) | ||
{ | ||
int f = isap(E[cur].endp, min(curf - totalf, E[cur].flow)); | ||
E[cur].flow -= f; | ||
E[cur ^ 1].flow += f; | ||
totalf += f; | ||
} | ||
if (V[E[cur].endp].dis < mindis) | ||
{ | ||
mindis = V[E[cur].endp].dis; | ||
} | ||
} | ||
} | ||
if (totalf == 0) | ||
{ | ||
if (--gap[V[u].dis] == 0) V[src].dis = nl + nr + 2; | ||
V[u].dis = mindis + 1; | ||
gap[V[u].dis]++; | ||
} | ||
return totalf; | ||
} | ||
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int max_flow() | ||
{ | ||
int res = 0; | ||
while (V[src].dis < nl + nr + 2) | ||
{ | ||
res += isap(src, INF); | ||
} | ||
return res; | ||
} | ||
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int main() | ||
{ | ||
int u, v; | ||
scanf("%d%d%d", &nl, &nr, &iec); | ||
for (int i = 0; i < iec; i++) | ||
{ | ||
scanf("%d%d", &u, &v); | ||
add_edge(u, v + nl, 1); | ||
add_edge(v + nl, u, 0); | ||
} | ||
for (int i = 1; i <= nl; i++) | ||
{ | ||
add_edge(src, i, 1); | ||
add_edge(i, src, 0); | ||
} | ||
for (int i = 1; i <= nr; i++) | ||
{ | ||
add_edge(i + nl, sink, 1); | ||
add_edge(sink, i + nl, 0); | ||
} | ||
init(); | ||
printf("%d\n", max_flow()); | ||
for (int i = 1; i <= nl; i++) | ||
{ | ||
bool has = false; | ||
for (int cur = V[i].first; cur != 0; cur = E[cur].next) | ||
{ | ||
if (E[cur].endp != src && E[cur].flow == 0) | ||
{ | ||
printf("%d ", E[cur].endp - nl); | ||
has = true; | ||
break; | ||
} | ||
} | ||
if (has == false) | ||
{ | ||
printf("0 "); | ||
} | ||
} | ||
return 0; | ||
} |