HDU 3488 Tour 最大完美匹配

有N个城市,M条街道,每条街道是单向的,现在要你设计多条路线覆盖所有的点,每条路线都是一个环,并且每个点仅能被一条路线覆盖且只经过一次

因为是圈,所以每个点的入度和出度应该都是1,故将一个点拆成两个点,入度点和出度点,可以想到，网络流的过程就是度数的流，因此可以建图跑最大匹配

#include 
#include 
#include 
#include 
#define rep(i, j, k) for(int i = j; i <= k; i++)
#define maxn 209
#define maxm 30009
#define inf 0x7fffffff

using namespace std;

int n, m, vis_x[maxn], vis_y[maxn], lx[maxn], ly[maxn], head[maxn], slack[maxn], link[maxn];
int value[maxm], to[maxm], nxt[maxm], Val[maxn][maxn], tot;

void add (int u, int v, int w)
{
    to[++tot] = v;
    nxt[tot] = head[u];
    head[u] = tot;
    value[tot] = w;
}

bool dfs (int x)
{
    vis_x[x] = 1;
    for (int i = head[x]; i; i = nxt[i])
        if (vis_y[to[i]] == 0)
        {
            int tmp = lx[x] + ly[to[i]] - Val[x][to[i]];
            if (tmp == 0)
            {
                vis_y[to[i]] = 1;
                if (link[to[i]] == -1 || dfs (link[ to[i] ]))
                {
                    link[to[i]] = x;
                    return 1;
                }
            }
            else
                slack[to[i]] = min (slack[to[i]], tmp);
        }
    return 0;
}

int km ()
{
    memset (link, -1, sizeof (link));
    memset (ly, 0, sizeof (ly));
    rep (x, 1, n)
    {
        lx[x] = -inf;
        for (int i = head[x]; i; i = nxt[i])
            lx[x] = max (lx[x], value[i]);
    }
    rep (x, 1, n)
    {
        rep (i, 1, n)
            slack[i] = inf;
        while (1)
        {
            memset (vis_x, 0, sizeof (vis_x));
            memset (vis_y, 0, sizeof (vis_y));
            if (dfs (x))
                break;
            int d = inf;
            rep (i, 1, n)
                if (!vis_y[i])
                    d = min (d, slack[i]);
            rep (i, 1, n)
                if (vis_y[i] == 0)
                    slack[i] -= d;
                else
                    ly[i] += d;
            rep (i, 1, n)
                if (vis_x[i])
                    lx[i] -= d;
        }
    }
    int ret = 0;
    rep (i, 1, n)
        if (link[i] != -1)
            ret -= Val[ link[i] ][i];
    return ret;
}

int main ()
{
    int ti;
    cin >> ti;
    while (ti--)
    {
        memset (head, 0, sizeof (head));
        memset (nxt, 0, sizeof (nxt));
        tot = 0;
        cin >> n >> m;
        rep (i, 1, n)
            rep (j, 1, m)
                Val[i][j] = -inf;
        rep (i, 1, m)
        {
            int u, v, w;
            scanf ("%d%d%d", &u, &v,&w);
            Val[u][v] = max (Val[u][v], -w);
            add (u, v, -w);
        }
        cout << km () << endl;
    }
    return 0;
}