题目

传送门

题解

第一步自然是缩点变成无向图。
然后就是裸的可相交最小路径覆盖。

不可相交最小路径覆盖

即用最少的路径覆盖图中所有点， 路径不可相交。

可用最小割解决。
考虑把原图中的每一个点看成一条单独的路径， 接着不断将路径合并， 形成更长的路径。
那么最后的路径数 = 总点数 - 合并次数\((\)没合并次减少一条路径\()\)， 要使路径数最小就要让合并次数最多。
可用二分图匹配计算。
把原图中的一个点\(u\)， 拆成\(u_1\)和\(u_2\)；把原图中的边\((u, v)\)， 改成\((u_1, v_2)\)。 那么就构成了一个二分图。在二分图中加一条匹配边等于一次合并， 最大匹配即为最大合并次数。

可相交最小路径覆盖

做法一：
即用最少的路径覆盖图中所有点， 路径可相交。

考虑把在上一个问题的算法直接搬过来。 显然是错的， 因为如果两个点的路径中的任意一个点有被覆盖过， 则两个点就会被无法被合并。
我们要消除这种影响。 换句话说， 只要\(A\)可到达\(B\)， 那么在二分图就要有一条\((A_1, B_2)\)的边。

然后在求二分图匹配， 即可得到正确的合并次数。

做法二：
最大费用最大流摸拟贪心。

代码

做法一：

#include <iostream>
#include <cstdlib>
#include <stack>
#include <cstdio>
#include <cstring>
#include <bitset>
#include <queue>using namespace std;const int N = 1210, M = 200010;const int INF = 0x3F3F3F3F;struct edge
{   int from, to, flow, cap;edge() { }edge(int _1, int _2, int _3, int _4) : from(_1), to(_2), flow(_3), cap(_4) { }
};namespace Graph
{   edge edges[M];int head[N], nxt[M], tot;inline void init(){   memset(head, -1, sizeof(head));tot = 0;}inline void add_edge(int x, int y){       edges[tot] = edge(x, y, 0, 0);nxt[tot] = head[x];head[x] = tot++;}
}namespace DAG
{   edge edges[M];int head[N], nxt[M], tot;inline void init(){   memset(head, -1, sizeof(head));tot = 0;}inline void add_edge(int x, int y){   edges[tot] = edge(x, y, 0, 0);nxt[tot] = head[x];head[x] = tot++;}
}struct Dinic
{   edge edges[N * N * 2];int head[2 * N], nxt[N * N * 2], tot;inline void init(){   memset(head, -1, sizeof(head));tot = 0;}inline void add_edge(int x, int y, int z){   edges[tot] = edge(x, y, 0, z);nxt[tot] = head[x];head[x] = tot++;edges[tot] = edge(y, x, 0, 0);nxt[tot] = head[y];head[y] = tot++;}int L, R;int s, t;int d[2 * N];bool bfs(){   memset(d, -1, sizeof(d));queue<int> q;q.push(s);d[s] = 0;while (!q.empty()){   int x = q.front(); q.pop();for (int i = head[x]; ~i; i = nxt[i]){   edge & e = edges[i];if (e.cap > e.flow && d[e.to] == -1){   d[e.to] = d[x] + 1;q.push(e.to);}}}return d[t] != -1;}int cur[2 * N];int dfs(int x, int a){   if (x == t || a == 0) return a;int f, flow = 0;for (int & i = cur[x]; ~i; i = nxt[i]){   edge & e = edges[i];if (d[e.to] == d[x] + 1 && (f = dfs(e.to, min(e.cap-e.flow, a))) > 0){   e.flow += f;edges[i^1].flow -= f;flow += f;a -= f;if (a == 0) break;}}return flow;}int maxflow(int _s, int _t){   s = _s, t = _t;int flow = 0;while (bfs()){   for (int i = L; i <= R; i++)cur[i] = head[i];flow += dfs(s, INF);}return flow;}
} dinic;int n, m;bool bo1[N], bo2[N], boo1[N], boo2[N];int dfn[N], low[N], dfs_clock;
int num[N], scc;
stack<int> st;bool boo[N], bo[N];void Tarjan(int x)
{   using namespace Graph;low[x] = dfn[x] = ++dfs_clock;st.push(x);for (int i = head[x]; ~i; i = nxt[i]){   edge & e = edges[i];if (!dfn[e.to]){   Tarjan(e.to);low[x] = min(low[x], low[e.to]);}else if (!num[e.to]) low[x] = min(low[x], dfn[e.to]);}if (low[x] == dfn[x]){   int v;++scc;do{   v = st.top(); st.pop();num[v] = scc;if (boo1[v]) bo1[scc] = 1;if (boo2[v]) bo2[scc] = 1;}while(v != x);}
}int out[N], in[N];int Ans;queue<int> q;int node[N], total;bitset<N> f[N];void solve()
{   using namespace DAG;for (int i = 1; i <= scc; i++)if (in[i] == 0 && !bo1[i]){   puts("no solution");return;}for (int i = 1; i <= scc; i++)if (out[i] == 0 && !bo2[i]){   puts("no solution");return;}total = 0;while (!q.empty()) q.pop();for (int i = 1; i <= scc; i++)if (in[i] == 0)q.push(i), node[++total] = i;while (!q.empty()){   int x = q.front(); q.pop();for (int i = head[x]; ~i; i = nxt[i]){   edge & e = edges[i];if ((--in[e.to]) == 0)q.push(e.to),node[++total] = e.to;}}int S = 2 * scc + 1, T = 2 * scc + 2;dinic.L = 0, dinic.R = T;dinic.init();for (int i = 1; i <= scc; i++)dinic.add_edge(S, i, 1);for (int i = 1; i <= scc; i++)dinic.add_edge(i+scc, T, 1);for (int i = total; i >= 1; i--){   int now = node[i];f[now][now-1] = 1;for (int j = head[now]; ~j; j = nxt[j]){   edge & e = edges[j];f[now] |= f[e.to];}for (int j = 1; j <= scc; j++)if (f[now][j-1] && j != now)dinic.add_edge(now, j+scc, 1);}for (int i = 1; i <= n; i++)f[i].reset();printf("%d\n", scc - dinic.maxflow(S, T));
}void build_DAG()
{   DAG::init();for (int i = 1; i <= n; i++){   for (int j = Graph::head[i]; ~j; j = Graph::nxt[j]){   edge & e = Graph::edges[j];if (num[i] != num[e.to])DAG::add_edge(num[i], num[e.to]),in[num[e.to]]++,out[num[i]]++;}}
}void clear()
{   memset(boo1, 0, sizeof(boo1));memset(boo2, 0, sizeof(boo2));memset(bo1, 0, sizeof(bo1));memset(bo2, 0, sizeof(bo2));memset(num, 0, sizeof(num));memset(dfn, 0, sizeof(dfn));memset(low, 0, sizeof(low));memset(in, 0, sizeof(in));memset(out, 0, sizeof(out));dfs_clock = 0;scc = 0;while (!st.empty()) st.pop();
}int main()
{   int T;scanf("%d", &T);while (T--){   int a, b;scanf("%d %d %d %d", &n, &m, &a, &b);clear();for (int i = 1; i <= a; i++){   int x;scanf("%d", &x);boo1[x] = 1;}for (int i = 1; i <= b; i++){   int x;scanf("%d", &x);boo2[x] = 1;}Ans = 0;Graph::init();for (int i = 1; i <= m; i++){   int x, y;scanf("%d %d", &x, &y);Graph::add_edge(x, y);}for (int i = 1; i <= n; i++)if (!dfn[i]) Tarjan(i);build_DAG();solve();}return 0;
}

做法二：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <algorithm>#include <queue>
#include <stack>using namespace std;const int N = 1010, M = 10010;const int INF = 0x3F3F3F3F;struct edge
{int from, to, flow, cap, dis;edge() { }edge(int _1, int _2, int _3, int _4, int _5) : from(_1), to(_2), flow(_3), cap(_4), dis(_5) { }
};namespace Graph
{edge edges[2*M];int head[N], nxt[2*M], tot;inline void init(){memset(head, -1, sizeof(head));tot = 0;}inline void add_edge(int x, int y){edges[tot] = edge(x, y, 0, 0, 0);nxt[tot] = head[x];head[x] = tot++;}
}struct MCMF
{edge edges[2 * N * N];int head[2*N], nxt[2 * N * N], tot;inline void init(){memset(head, -1, sizeof(head));tot = 0;}inline void add_edge(int x, int y, int z, int k){edges[tot] = edge(x, y, 0, z, k);nxt[tot] = head[x];head[x] = tot++;edges[tot] = edge(y, x, 0, 0, -k);nxt[tot] = head[y];head[y] = tot++;}int s, t;int L, R;int dist[2*N]; bool inq[2*N];int a[2*N], p[2*N];bool SPFA(){for (int i = L; i <= R; i++)dist[i] = -INF, inq[i] = 0;queue <int> q;q.push(s);inq[s] = 1;dist[s] = 0;a[s] = INF;p[s] = 0;while (!q.empty()){int x = q.front(); q.pop();inq[x] = 0;for (int i = head[x]; ~i; i = nxt[i]){edge & e = edges[i];if (e.cap > e.flow && dist[e.to] < dist[x] + e.dis){dist[e.to] = dist[x] + e.dis;a[e.to] = min(a[x], e.cap - e.flow);p[e.to] = i;if (!inq[e.to]) q.push(e.to), inq[e.to] = 1;}}}if (dist[t] <= 0) return 0;for (int i = t; i != s; i = edges[p[i]].from){edges[p[i]].flow += a[t];edges[p[i]^1].flow -= a[t];}return 1;}int calc(int _s, int _t){int Ans = 0;s = _s, t = _t;while (SPFA()) Ans++;return Ans;}
} maxcostminflow;int n, m, a, b;bool type1[N], type2[N];bool bo1[N], bo2[N];int low[N], dfn[N], dfs_clock;
int bel[N], cnt;
stack <int> stk;void Tarjan(int x)
{using namespace Graph;low[x] = dfn[x] = ++dfs_clock;stk.push(x);for (int i = head[x]; ~i; i = nxt[i]){edge & e = edges[i];if (!dfn[e.to]){Tarjan(e.to);low[x] = min(low[x], low[e.to]);}else if (!bel[e.to]) low[x] = min(low[x], dfn[e.to]);}if (low[x] == dfn[x]){int v = 0;cnt++;do{v = stk.top(); stk.pop();bel[v] = cnt;if (type1[v]) bo1[cnt] = 1;if (type2[v]) bo2[cnt] = 1;}while (v != x);}
}bool g[N][N];int deg1[N], deg2[N];void solve()
{memset(dfn, 0, sizeof(dfn));memset(bel, 0, sizeof(bel));memset(g, 0, sizeof(g));memset(bo1, 0, sizeof(bo1));memset(bo2, 0, sizeof(bo2));cnt = 0;dfs_clock = 0;for (int i = 1; i <= n; i++)if (!dfn[i]) Tarjan(i);for (int i = 0; i < Graph::tot; i++){edge & e = Graph::edges[i];if (bel[e.from] != bel[e.to]) g[bel[e.from]][bel[e.to]] = 1;}memset(deg1, 0, sizeof(deg1));memset(deg2, 0, sizeof(deg2));for (int i = 1; i <= cnt; i++)for (int j = 1; j <= cnt; j++)if (i != j && g[i][j])deg2[i]++, deg1[j]++;for (int i = 1; i <= cnt; i++)if (!deg1[i] && !bo1[i]) return void(puts("no solution"));for (int i = 1; i <= cnt; i++)if (!deg2[i] && !bo2[i]) return void(puts("no solution"));maxcostminflow.init();maxcostminflow.L = 0, maxcostminflow.R = 2 * cnt + 2;   int S = cnt + cnt + 1, T = cnt + cnt + 2;for (int i = 1; i <= cnt; i++)maxcostminflow.add_edge(S, i, INF, 0);for (int i = 1; i <= cnt; i++)maxcostminflow.add_edge(i+cnt, T, INF, 0);for (int i = 1; i <= cnt; i++){maxcostminflow.add_edge(i, i+cnt, 1, 1);maxcostminflow.add_edge(i, i+cnt, INF, 0);}for (int i = 1; i <= cnt; i++)for (int j = 1; j <= n; j++)if (g[i][j])maxcostminflow.add_edge(i+cnt, j, INF, 0);printf("%d\n", maxcostminflow.calc(S, T));
}int main()
{int T;scanf("%d", &T);while (T--){memset(type1, 0, sizeof(type1));memset(type2, 0, sizeof(type2));scanf("%d %d %d %d", &n, &m, &a, &b);Graph::init();for (int i = 1; i <= a; i++){int x;scanf("%d", &x);type1[x] = 1;}for (int i = 1; i <= b; i++){int x;scanf("%d", &x);type2[x] = 1;}for (int i = 1; i <= m; i++){int x, y;scanf("%d %d", &x, &y);Graph::add_edge(x, y);}solve();}return 0;
}