[hdu1532]最大流
裸最大流，求最大流一般步骤如下：
（1）所有正向边权初始化为容量，反向边权初始化为0
（2）找增广路
（3）找到则进入（4），否则得到最大流并退出
（4） 增广路上所有边减去最小边权，相应的方向边加上最小边权，然后返回（2）
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#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <ctime>
#include <deque>
#include <queue>
using
namespace
std;
struct
MaxFlow {
private
:
const
static
int
maxn = 2e2 + 7;
struct
Edge {
int
u, v, w;
Edge(
int
u = 0,
int
v = 0,
int
w = 0): u(u), v(v), w(w) {}
};
vector<vector<
int
> > G;
vector<Edge> E;
int
S, T, maxflow;
bool
vis[maxn];
int
Q[maxn], fa[maxn], Y[maxn], head, tail, flow[maxn];
public
:
void
init(
int
s,
int
t,
int
n) {
G.clear();
G.resize(n + 2);
E.clear();
S = s;
T = t;
maxflow = 0;
}
void
add(
int
u,
int
v,
int
w) {
E.push_back(Edge(u, v, w));
E.push_back(Edge(v, u, 0));
int
sz = E.size();
G[u].push_back(sz - 2);
G[v].push_back(sz - 1);
}
bool
bfs(
int
src) {
head = tail = 0;
memset
(vis, 0,
sizeof
(vis));
Q[tail ++] = src;
flow[0] = 0x7fffffff;
vis[src] =
true
;
while
(head < tail) {
int
node = Q[head ++];
if
(node == T) {
maxflow += flow[head - 1];
int
p = head - 1;
while
(p) {
E[Y[p]].w -= flow[head - 1];
E[Y[p] ^ 1].w += flow[head - 1];
p = fa[p];
}
return
true
;
}
for
(
int
i = 0; i < G[node].size(); i ++) {
int
e = G[node][i];
if
(!vis[E[e].v] && E[e].w) {
vis[E[e].v] =
true
;
fa[tail] = head - 1;
Y[tail] = e;
flow[tail] = min(flow[head - 1], E[e].w);
Q[tail ++] = E[e].v;
}
}
}
return
false
;
}
int
solve() {
while
(bfs(S));
return
maxflow;
}
} ;
MaxFlow solver;
int
main() {
#ifndef ONLINE_JUDGE
freopen
(
"in.txt"
,
"r"
, stdin);
#endif // ONLINE_JUDGE
int
n, m;
while
(cin >> m >> n) {
solver.init(1, n, n);
for
(
int
i = 0; i < m; i ++) {
int
u, v, w;
scanf
(
"%d%d%d"
, &u, &v, &w);
solver.add(u, v, w);
}
cout << solver.solve() << endl;
}
return
0;
}
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