1.樸素KM 時間複雜度：$O(n^4)$

演算法過程：

1. 初始化可行頂標的值 (設定lx,ly的初始值)
2. 用匈牙利演算法尋找相等子圖的完備匹配
3. 若未找到增廣路則修改可行頂標的值
4. 重複(2)(3)直到找到相等子圖的完備匹配為止

板子：

#pragma GCC optimize(3)
#pragma GCC optimize(2)
#include <iostream>
#include <string>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include 
 
<cmath>
#include <queue>
#include <functional>
#include <vector>
#include <map>
#include <set>
#include <stack>
#include <fstream>
#include <sstream>
#include <unordered_map>
#define FT(a, b) memset(a, b, sizeof(a))
#define FAT(a) memset(a, 0, sizeof(a))
using
 
 namespace std;
typedef long long ll;
const int N = 5e2 + 10;
const int M = 1e5 + 100;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int n, m, k, x, y, t;
int g[N][N], lx[N], ly[N], visx[N], visy[N], match[N];
bool findpath(int x)
{
    for (int i = 1; i < n + 1; ++i)
    {
        if (visy[i] || lx[x] + ly[i] != g[x][i])
            
 
continue;
        visy[i] = 1;
        if (match[i] == -1 || findpath(match[i]))
        {
            match[i] = x;
            return 1;
        }
    }
    return 0;
}
ll km()
{
    ll ans = 0;
    for (int i = 1; i < n + 1; ++i)
    {
        while (1)
        {
            FAT(visx), FAT(visy);
            int delta = INF;
            if (findpath(i))
                break;
            else
            {
                for (int i = 1; i < n + 1; ++i)
                    if (visx[i])
                        for (int j = 1; j < n + 1; ++j)
                            if (visy[j])
                                delta = min(delta, lx[i] + ly[j] - g[i][j]);
                for (int i = 1; i < n + 1; ++i)
                {
                    if (visx[i])
                        lx[i] -= delta;
                    if (visy[i])
                        ly[i] += delta;
                }
            }
        }
    }
    for (int i = 1; i < n + 1; ++i)
        ans += lx[i];
    return ans;
}
void solve()
{
    scanf("%d%d", &n, &m);
    FAT(lx), FAT(ly);
    FT(g, INF), FT(match, -1);
    for (int i = 0; i < m; ++i)
    {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        g[a][b] = c;
        lx[a] = max(lx[a], c);
    }
    printf("%lld\n", km());
    for (int i = 1; i < n + 1; ++i)
        printf("%d ", match[i]);
    puts("");
}

int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("/home/wjy/code/in.txt", "r", stdin);
    // freopen("/home/wjy/code/ans1.txt","w", stdout);
#endif
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int T = 1;
    // scanf("%d", &T);
    // cin >> T;
    // time_t begin, end;
    // begin = clock();
    while (T--)
    {
        solve();
    }
    // end = clock();
    // double ret = double(end - begin) / CLOCKS_PER_SEC;
    return 0;
}