Source: Light OJ 1406 Assassin's Creed
Directed Graph sends the least people to all cities and no one can walk anywhere else.
Ideas: the minimum number of people who can finish the whole graph is obviously the problem of least path coverage. Here there may be loops, so we need to shrink the point. However, we can see that a strongly connected component may have to be split into N at most 15, so the state is compressed.
Divide the full graph into sub-States. Each sub-state is scaled down to obtain the minimum path coverage. This solves the issue of splitting strongly connected components. Finally, the state is compressed and the DP is used to solve the optimal value.
#include <cstdio>#include <cstring>#include <vector>#include <algorithm>#include <stack>using namespace std;const int maxn = 16;vector <int> G[maxn], G2[maxn];int dp[1<<maxn];int pre[maxn], low[maxn], sccno[maxn];int clock, scc_cnt;int n, m;stack <int> S;int a[maxn][maxn];int b[maxn][maxn];void dfs(int u, int x){pre[u] = low[u] = ++clock;S.push(u);for(int i = 0; i < G[u].size(); i++){int v = G[u][i];if(!(x&(1<<v)))continue;if(!pre[v]){dfs(v, x);low[u] = min(low[u], low[v]); }else if(!sccno[v]){low[u] = min(low[u], pre[v]);}}if(pre[u] == low[u]){scc_cnt++;while(1){int x = S.top(); S.pop();sccno[x] = scc_cnt;if(x == u)break;}}}int find_scc(int x){memset(sccno, 0, sizeof(sccno));memset(pre, 0, sizeof(pre));scc_cnt = 0, clock = 0;for(int i = 0; i < n; i++){if(x&(1<<i) && !pre[i])dfs(i, x);}return scc_cnt;}int y[maxn];bool vis[maxn];bool xyl(int u){for(int i = 0; i < G2[u].size(); i++){int v = G2[u][i];if(vis[v])continue;vis[v] = true;if(y[v] == -1 || xyl(y[v])){y[v] = u;return true;}}return false;}int match(){int ans = 0;memset(y, -1, sizeof(y));for(int i = 1; i <= scc_cnt; i++){memset(vis, false, sizeof(vis));if(xyl(i))ans++;}return scc_cnt-ans;}int main(){int cas = 1;int T;scanf("%d", &T);while(T--){scanf("%d %d", &n, &m);for(int i = 0; i < n; i++)G[i].clear();memset(a, 0, sizeof(a));while(m--){int u, v;scanf("%d %d", &u, &v);u--;v--;G[u].push_back(v);a[u][v] = 1;}dp[0] = 0;//puts("sdf");for(int i = 1; i < (1<<n); i++){//memset(b, 0, sizeof(b));find_scc(i);for(int j = 0; j <= n; j++)G2[j].clear();for(int j = 0; j < n; j++)for(int k = 0; k < n; k++)if(a[j][k] && sccno[j] && sccno[k] && sccno[j] != sccno[k])G2[sccno[j]].push_back(sccno[k]);dp[i] = match();}//puts("sdf");for(int s = 1; s < (1<<n); s++){for(int i = s; i; i = s&(i-1)){dp[s] = min(dp[s], dp[s^i] + dp[i]);}}printf("Case %d: %d\n", cas++, dp[(1<<n)-1]);}return 0;}