您的位置:首页 > 理论基础 > 计算机网络

图论 网络流 HDU 1532 最大流

2016-10-13 00:48 330 查看
题意:最大流,n边,m点,求1点到m点 最大流。

题解:最大流模板

#include <iostream>
#include <cstdio>
#include <math.h>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
const int MAXNODE = 105 * 2;
const int MAXEDGE = 100005;
typedef int Type;
const Type INF = 0x3f3f3f3f;

struct Edge {
int u, v;
Type cap, flow;
Edge() {}
Edge(int u, int v, Type cap, Type flow) {
this->u = u;
this->v = v;
this->cap = cap;
this->flow = flow;
}
};

struct Dinic {
int n, m, s, t;
Edge edges[MAXEDGE];
int first[MAXNODE];
int next[MAXEDGE];
bool vis[MAXNODE];
Type d[MAXNODE];
int cur[MAXNODE];
vector<int> cut;

void init(int n) {
this->n = n;
memset(first, -1, sizeof(first));
m = 0;
}
void add_Edge(int u, int v, Type cap) {
edges[m] = Edge(u, v, cap, 0);
next[m] = first[u];
first[u] = m++;
edges[m] = Edge(v, u, 0, 0);
next[m] = first[v];
first[v] = m++;
}

bool bfs() {
memset(vis, false, sizeof(vis));
queue<int> Q;
Q.push(s);
d[s] = 0;
vis[s] = true;
while (!Q.empty()) {
int u = Q.front(); Q.pop();
for (int i = first[u]; i != -1; i = next[i]) {
Edge& e = edges[i];
if (!vis[e.v] && e.cap > e.flow) {
vis[e.v] = true;
d[e.v] = d[u] + 1;
Q.push(e.v);
}
}
}
return vis[t];
}

Type dfs(int u, Type a) {
if (u == t || a == 0) return a;
Type flow = 0, f;
for (int &i = cur[u]; i != -1; i = next[i]) {
Edge& e = edges[i];
if (d[u] + 1 == d[e.v] && (f = dfs(e.v, min(a, e.cap - e.flow))) > 0) {
e.flow += f;
edges[i^1].flow -= f;
flow += f;
a -= f;
if (a == 0) break;
}
}
return flow;
}

Type Maxflow(int s, int t) {
this->s = s; this->t = t;
Type flow = 0;
while (bfs()) {
for (int i = 0; i < n; i++)
cur[i] = first[i];
flow += dfs(s, INF);
}
return flow;
}

void MinCut() {
cut.clear();
for (int i = 0; i < m; i += 2) {
if (vis[edges[i].u] && !vis[edges[i].v])
cut.push_back(i);
}
}
} gao;
int main()
{
int n,m;
while(~scanf("%d%d",&n,&m)){//n 边 m点
int a,b,c;
gao.init(n); //边
for(int i=1;i<=n;i++){
scanf("%d%d%d",&a,&b,&c);
gao.add_Edge(a,b,c);
}
printf("%d\n",gao.Maxflow(1,m));

}
return 0;
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: