[SWUST1738] 最小路径覆盖问题（最大流，最小路径覆盖）

题目链接：https://www.oj.swust.edu.cn/problem/show/1738

把每一个数拆成两个点，建图跑最大流，结论是满足最小路径覆盖的路径数=总点数-最小割，即总点数-最大流。

打印路径dfs一下，非递归的时候PE了，但是这个OJ会报WA，哎。

#include <bits/stdc++.h>
using namespace std;

typedef struct Edge {
    int u, v, w, next;
}Edge;

const int inf = 0x7f7f7f7f;
const int maxn = 9090;

int cnt, dhead[maxn];
int cur[maxn], dd[maxn];
Edge dedge[maxn<<3];
int S, T, N;

void init() {
    memset(dhead, -1, sizeof(dhead));
    for(int i = 0; i < maxn; i++) dedge[i].next = -1;
    S = 0; cnt = 0;
}

void adde(int u, int v, int w, int c1=0) {
    dedge[cnt].u = u; dedge[cnt].v = v; dedge[cnt].w = w; 
    dedge[cnt].next = dhead[u]; dhead[u] = cnt++;
    dedge[cnt].u = v; dedge[cnt].v = u; dedge[cnt].w = c1; 
    dedge[cnt].next = dhead[v]; dhead[v] = cnt++;
}

bool bfs(int s, int t, int n) {
    queue<int> q;
    for(int i = 0; i < n; i++) dd[i] = inf;
    dd[s] = 0;
    q.push(s);
    while(!q.empty()) {
        int u = q.front(); q.pop();
        for(int i = dhead[u]; ~i; i = dedge[i].next) {
            if(dd[dedge[i].v] > dd[u] + 1 && dedge[i].w > 0) {
                dd[dedge[i].v] = dd[u] + 1;
                if(dedge[i].v == t) return 1;
                q.push(dedge[i].v);
            }
        }
    }
    return 0;
}

int dinic(int s, int t, int n) {
    int st[maxn], top;
    int u;
    int flow = 0;
    while(bfs(s, t, n)) {
        for(int i = 0; i < n; i++) cur[i] = dhead[i];
        u = s; top = 0;
        while(cur[s] != -1) {
            if(u == t) {
                int tp = inf;
                for(int i = top - 1; i >= 0; i--) {
                    tp = min(tp, dedge[st[i]].w);
                }
                flow += tp;
                for(int i = top - 1; i >= 0; i--) {
                    dedge[st[i]].w -= tp;
                    dedge[st[i] ^ 1].w += tp;
                    if(dedge[st[i]].w == 0) top = i;
                }
                u = dedge[st[top]].u;
            }
            else if(cur[u] != -1 && dedge[cur[u]].w > 0 && dd[u] + 1 == dd[dedge[cur[u]].v]) {
                st[top++] = cur[u];
                u = dedge[cur[u]].v;
            }
            else {
                while(u != s && cur[u] == -1) {
                    u = dedge[st[--top]].u;
                }
                cur[u] = dedge[cur[u]].next;
            }
        }
    }
    return flow;
}

int n, m;
vector<int> path;
bool vis[maxn];

void dfs(int u) {
    vis[u] = 1;
    path.push_back(u);
    for(int j = dhead[u]; ~j; j=dedge[j].next) {
        if(vis[dedge[j].v]) continue;
        if(dedge[j].w || !dedge[j].v) continue;
        dfs(dedge[j].v-n);
    }
}

int main() {
    // freopen("in", "r", stdin);
    int u, v, w;
    while(~scanf("%d%d",&n,&m)) {
        init();
        S = 0; T = 2 * n + 1; N = T + 1;
        for(int i = 1; i <= n; i++) adde(S, i, 1);
        for(int i = 1; i <= n; i++) adde(n+i, T, 1);
        for(int i = 0; i < m; i++) {
            scanf("%d%d",&u,&v);
            adde(u, n+v, 1);
        }
        int ret = dinic(S, T, N);
        memset(vis, 0, sizeof(vis));
        for(int i = 1; i <= n; i++) {
            if(vis[i]) continue;
            path.clear();
            dfs(i);
            for(int j = 0; j < path.size(); j++) {
                printf("%d%c", path[j], j==path.size()-1?'
':' ');
            }
        }
        printf("%d
", n - ret);
    }
    return 0;
}