Ford-Fulkerson算法 java实现

/*

for each edge(u,v) 属于 G,E

(u,v).f = 0;

while there exists a path p from s to t in the residual network Gf

cf(p) = min{cf(u,v):(u,v) is in p};

for each edge(u,v) in p

if(u,v)属于E

(u,v).f += cf(p)

else

(u,v).f -= cf(p);

*/

用数组实现栈来进行深度遍历的地方(dfs函数)需要尤其注意。

import java.util.*;
class Ford_Fulkerson
{
static int[][] res;	//残留矩阵
static int[] pre;
static boolean[] used;
static int[] stack; //用于dfs，底部为0，顶部为top
static int ver;
static int edge;

static int ford_Fulkerson(int s,int t){
//res默认初始化为0
int cfp = Integer.MAX_VALUE;
int maxflow = 0;
while(dfs(s,t)){
int d = Integer.MAX_VALUE;
for(int i = t; i != s; i = pre[i])
d = Math.min(d, res[pre[i]][i]);
for(int i=t; i != s; i = pre[i])
{
res[pre[i]][i] -= d;
res[i][pre[i]] += d;
}
maxflow += d;
}
return maxflow;
}

static boolean dfs(int s ,int t){
for(int i=0;i<=ver;i++) used[i] = false;
used[s] = true;
int top = 0;
stack[0] = s; top++;
int cur = s,i = 1;
int[] rec = new int[ver+5]; //记录第i个节点遍历到的位置
boolean flag = false;
//采用非递归的形式，找到一条增广路径后立刻终止
while(top != 0){ //stack 不为空
flag = false;
cur = stack[top-1]; //相当于peek操作
for(i=rec[cur]+1;i <= ver;i++){
//取与其相邻的并且没有访问过的值大于0的点入栈
if(used[i]|| res[cur][i] <= 0) continue;
flag = true;
pre[i] = cur;
used[i] = true;
if(i == t) return true; //找到目标节点，返回true
stack[top++] = i;	//入栈
rec[cur] = i;
break;
}
//////////////////////////////////////////////////////////////////
//////////////////////////////////////////////////////////////////
if(!flag) {used[stack[--top]] = false;}//回溯过程恢复设置,并pop
}
return false;
}

static void init(){
res = new int[ver+5][ver+5];
used = new boolean[ver+5];
pre = new int[ver+5];
stack = new int[ver+5];
}

static void input(){
Scanner cin = new Scanner(System.in);
System.out.println("请输入 点数 边数");
ver = cin.nextInt();
edge = cin.nextInt();
int s,e,w;
init();
System.out.println("起点 终点 权值");
for(int i=0;i<edge;i++){
s = cin.nextInt();
e = cin.nextInt();
w = cin.nextInt();
res[s][e] = w;
}
}

public static void main(String[] args){
input();
System.out.println(ford_Fulkerson(1,4));
}
}