2022-06-11：注意本文件中，graph不是邻接矩阵的含义，而是一个二部图。 在长度为N的邻接矩阵matrix中，所有的点有N个，matrix[i][j]表示点i到点j的距离或者权重， 而在二部

Posted 2022-06-15 福大大架构师每日一题

2022-06-11：注意本文件中，graph不是邻接矩阵的含义，而是一个二部图。
在长度为N的邻接矩阵matrix中，所有的点有N个，matrix[i][j]表示点i到点j的距离或者权重，
而在二部图graph中，所有的点有2*N个，行所对应的点有N个，列所对应的点有N个。
而且认为，行所对应的点之间是没有路径的，列所对应的点之间也是没有路径的！

答案2022-06-11：

km算法。

代码用rust编写。代码如下：

use rand::Rng;
fn main() 
    let n: i32 = 10;
    let v: i32 = 20;
    let test_time: i32 = 10;
    println!("测试开始");
    for _ in 0..test_time 
        let mut graph = random_graph(n, v);
        let ans1 = right(&mut graph);
        let ans2 = km(&mut graph);
        if ans1 != ans2 
            println!("出错了!");
            println!("ans1 = ", ans1);
            println!("ans2 = ", ans2);
            println!("===============");
        
    
    println!("测试结束");


// 暴力解
fn right(graph: &mut Vec<Vec<i32>>) -> i32 
    let N = graph.len() as i32;
    let mut to: Vec<i32> = vec![];
    for _ in 0..N 
        to.push(1);
    
    return process(0, &mut to, graph);


fn process(from: i32, to: &mut Vec<i32>, graph: &mut Vec<Vec<i32>>) -> i32 
    if from == graph.len() as i32 
        return 0;
    
    let mut ans = 0;
    for i in 0..to.len() as i32 
        if to[i as usize] == 1 
            to[i as usize] = 0;
            ans = get_max(
                ans,
                graph[from as usize][i as usize] + process(from + 1, to, graph),
            );
            to[i as usize] = 1;
        
    
    return ans;


fn get_max<T: Clone + Copy + std::cmp::PartialOrd>(a: T, b: T) -> T 
    if a > b 
        a
     else 
        b
    


fn get_min<T: Clone + Copy + std::cmp::PartialOrd>(a: T, b: T) -> T 
    if a < b 
        a
     else 
        b
    


fn km(graph: &mut Vec<Vec<i32>>) -> i32 
    let N = graph.len() as i32;
    let mut match0: Vec<i32> = vec![];
    let mut lx: Vec<i32> = vec![];
    let mut ly: Vec<i32> = vec![];
    // dfs过程中，碰过的点！
    let mut x: Vec<bool> = vec![];
    let mut y: Vec<bool> = vec![];
    // 降低的预期！
    // 公主上，打一个，降低预期的值，只维持最小！
    let mut slack: Vec<i32> = vec![];
    let mut falsev: Vec<bool> = vec![];
    for _ in 0..N 
        match0.push(0);
        lx.push(0);
        ly.push(0);
        x.push(false);
        y.push(false);
        slack.push(0);
        falsev.push(false);
    
    let invalid = 2147483647;
    for i in 0..N 
        match0[i as usize] = -1;
        lx[i as usize] = -invalid;
        for j in 0..N 
            lx[i as usize] = get_max(lx[i as usize], graph[i as usize][j as usize]);
        
        ly[i as usize] = 0;
    
    for from in 0..N 
        for i in 0..N 
            slack[i as usize] = invalid;
        
        x = falsev.clone();
        y = falsev.clone();
        // dfs() : from王子，能不能不降预期，匹配成功！
        // 能：dfs返回true！
        // 不能：dfs返回false！
        while !dfs(
            from,
            &mut x,
            &mut y,
            &mut lx,
            &mut ly,
            &mut match0,
            &mut slack,
            graph,
        ) 
            // 刚才的dfs，失败了！
            // 需要拿到，公主的slack里面，预期下降幅度的最小值！
            let mut d = invalid;
            for i in 0..N 
                if !y[i as usize] && slack[i as usize] < d 
                    d = slack[i as usize];
                
            
            // 按照最小预期来调整预期
            for i in 0..N 
                if x[i as usize] 
                    lx[i as usize] = lx[i as usize] - d;
                
                if y[i as usize] 
                    ly[i as usize] = ly[i as usize] + d;
                
            
            x = falsev.clone();
            y = falsev.clone();
            // 然后回到while里，再次尝试
        
    
    let mut ans = 0;
    for i in 0..N 
        ans += lx[i as usize] + ly[i as usize];
    
    return ans;


// from, 当前的王子
// x，王子碰没碰过
// y, 公主碰没碰过
// lx，所有王子的预期
// ly, 所有公主的预期
// match，所有公主，之前的分配，之前的爷们！
// slack，连过，但没允许的公主，最小下降的幅度
// map，报价，所有王子对公主的报价
// 返回，from号王子，不降预期能不能配成！
fn dfs(
    from: i32,
    x: &mut Vec<bool>,
    y: &mut Vec<bool>,
    lx: &mut Vec<i32>,
    ly: &mut Vec<i32>,
    match0: &mut Vec<i32>,
    slack: &mut Vec<i32>,
    map: &mut Vec<Vec<i32>>,
) -> bool 
    let N = map.len() as i32;
    x[from as usize] = true;
    for to in 0..N 
        if !y[to as usize] 
            // 只有没dfs过的公主，才会去尝试
            let d = lx[from as usize] + ly[to as usize] - map[from as usize][to as usize];
            if d != 0 
                // 如果当前的路不符合预期，更新公主的slack值
                slack[to as usize] = get_min(slack[to as usize], d);
             else 
                // 如果当前的路符合预期，尝试直接拿下，或者抢夺让之前的安排倒腾去
                y[to as usize] = true;
                if match0[to as usize] == -1
                    || dfs(match0[to as usize], x, y, lx, ly, match0, slack, map)
                
                    match0[to as usize] = from;
                    return true;
                
            
        
    
    return false;


// 为了测试
fn random_graph(N: i32, V: i32) -> Vec<Vec<i32>> 
    let mut graph: Vec<Vec<i32>> = vec![];
    for i in 0..N 
        graph.push(vec![]);
        for _ in 0..N 
            graph[i as usize].push(0);
        
    
    for i in 0..N 
        for j in i + 1..N 
            let num = rand::thread_rng().gen_range(0, V);
            graph[i as usize][j as usize] = num;
            graph[j as usize][i as usize] = num;
        
    
    return graph;

执行结果如下：

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左神java代码