网络流最小路径覆盖

思路：

http://blog.csdn.net/tramp_1/article/details/52742572

每个点x拆成两个点x和x'，分别表示x作为前驱和作为后继。若原图中x和y有边，向x和y'加一条有向边。如此构成二分图，记此二分图中作为前驱的节点集合为A,作为后继的节点集合为B。跑最大匹配，没有匹配的点的个数（n-最大匹配数）就是需要的最少的路径条数。正确性：二分匹配可以保证每个点顶多只有一个前驱，并且顶多只有一个后继，也就保证了每个点在且仅在一条路径中。此外，在二分图中，A中每个没有匹配的顶点对应了一条路径的终点。（类似地，B中每一个没有匹配的顶点对应了一条路径的起点。）最大二分匹配可以保证A（或B）中没有匹配的点的数量最少，亦即路径条数最少。

实现：

#include <bits/stdc++.h>

#define N (1000 + 2)
#define M (N * N + 4 * N)

typedef long long LL;

using namespace std;

struct edge 
{
    int v, cap, next;
};
edge e[M];

int head[N], level[N], cur[N];
int num_of_edges;

/*
 * When there are multiple test sets, you need to re-initialize before each
 */
void dinic_init(void) 
{
    num_of_edges = 0;
    memset(head, -1, sizeof(head));
    return;
}

int add_edge(int u, int v, int c1, int c2) 
{
    int& i = num_of_edges;

    assert(c1 >= 0 && c2 >= 0 && c1 + c2 >= 0); // check for possibility of overflow
    e[i].v = v;
    e[i].cap = c1;
    e[i].next = head[u];
    head[u] = i++;

    e[i].v = u;
    e[i].cap = c2;
    e[i].next = head[v];
    head[v] = i++;
    return i;
}

void print_graph(int n) 
{
    for (int u=0; u<n; u++) 
    {
        printf("%d: ", u);
        for (int i=head[u]; i>=0; i=e[i].next) 
        {
            printf("%d(%d)", e[i].v, e[i].cap);
        }
        printf("
");
    }
    return;
}

/*
 * Find all augmentation paths in the current level graph
 * This is the recursive version
 */
int dfs(int u, int t, int bn) 
{
    if (u == t) return bn;
    int left = bn;
    for (int &i=cur[u]; i>=0; i=e[i].next) 
    {
        int v = e[i].v;
        int c = e[i].cap;
        if (c > 0 && level[u]+1 == level[v]) 
        {
            int flow = dfs(v, t, min(left, c));
            if (flow > 0) 
            {
                e[i].cap -= flow;
                e[i^1].cap += flow;
                cur[u] = i;
                left -= flow;
                if (!left) break;
            }
        }
    }
    if (left > 0) level[u] = 0;
    return bn - left;
}

bool bfs(int s, int t) 
{
    memset(level, 0, sizeof(level));
    level[s] = 1;
    queue<int> q;
    q.push(s);
    while (!q.empty()) 
    {
        int u = q.front();
        q.pop();
        if (u == t) return true;
        for (int i=head[u]; i>=0; i=e[i].next) 
        {
            int v = e[i].v;
            if (!level[v] && e[i].cap > 0) 
            {
                level[v] = level[u]+1;
                q.push(v);
            }
        }
    }
    return false;
}

LL dinic(int s, int t) 
{
    LL max_flow = 0;

    while (bfs(s, t)) 
    {
        memcpy(cur, head, sizeof(head));
        max_flow += dfs(s, t, INT_MAX);
    }
    return max_flow;
}

int main() 
{
    dinic_init();
    int m, n, x, y;
    cin >> m >> n;
    for (int i = 0; i < m; i++)
    {
        cin >> x >> y;
        add_edge(x, y + n, 1, 0);
    }
    for (int i = 1; i <= n; i++)
    {
        add_edge(0, i, 1, 0);
        add_edge(i + n, 2 * n + 1, 1, 0);
    }
    int ans = dinic(0, 2 * n + 1);
    cout << n - ans << endl;
    return 0;
}