【题目链接】
发现每条边最少走一次,最大走inf次,那么建图就出来了。
从源点向每个点连边,容量为inf。
原图中每条边的下界为1,上界为inf。
每个点向汇点连边,容量为inf。
跑有上下界的最小流就行了。
最小流写法看了【POPOQQQ的题解】
/* Pigonometry */ #include <cstdio> #include <algorithm> using namespace std; const int maxn = 105, maxm = 50005, maxq = 10000, inf = 0x3f3f3f3f; int n, head[maxn], cur[maxn], cnt, du[maxn], depth[maxn], bg, ed, S, T, q[maxq]; struct _edge { int v, w, next; } g[maxm << 1]; inline int iread() { int f = 1, x = 0; char ch = getchar(); for(; ch < '0' || ch > '9'; ch = getchar()) f = ch == '-' ? -1 : 1; for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0'; return f * x; } inline void add(int u, int v, int w) { g[cnt] = (_edge){v, w, head[u]}; head[u] = cnt++; } inline void link(int u, int v, int w) { add(u, v, w); add(v, u, 0); } inline bool bfs() { for(int i = 0; i <= T; i++) depth[i] = -1; int h = 0, t = 0, u, i; depth[q[t++] = bg] = 1; while(h != t) for(i = head[u = q[h++]]; ~i; i = g[i].next) if(g[i].w && !~depth[g[i].v]) { depth[g[i].v] = depth[u] + 1; if(g[i].v == ed) return 1; q[t++] = g[i].v; } return 0; } inline int dfs(int x, int flow) { if(x == ed) return flow; int left = flow; for(int i = cur[x]; ~i; i = g[i].next) if(g[i].w && depth[g[i].v] == depth[x] + 1) { int tmp = dfs(g[i].v, min(left, g[i].w)); left -= tmp; g[i].w -= tmp; g[i ^ 1].w += tmp; if(g[i].w) cur[x] = i; if(!left) return flow; } if(left == flow) depth[x] = -1; return flow - left; } int main() { n = iread(); bg = 0; ed = n + 1; S = n + 2; T = n + 3; for(int i = 0; i <= T; i++) head[i] = -1; cnt = 0; for(int i = 1; i <= n; i++) for(int x = iread(); x; x--) { int v = iread(); du[i]--; du[v]++; link(i, v, inf); } for(int i = 1; i <= n; i++) { link(S, i, inf); link(i, T, inf); if(du[i] > 0) link(bg, i, du[i]); if(du[i] < 0) link(i, ed, -du[i]); } link(T, S, inf); int ans = 0; while(bfs()) { for(int i = 0; i <= T; i++) cur[i] = head[i]; dfs(bg, inf); } for(int i = head[bg]; ~i; i = g[i].next) g[i].w = g[i ^ 1].w = 0; for(int i = head[ed]; ~i; i = g[i].next) g[i].w = g[i ^ 1].w = 0; for(int i = head[T]; ~i; i = g[i].next) if(g[i].v == S) { ans += g[i ^ 1].w; g[i].w = g[i ^ 1].w = 0; break; } link(bg, T, inf); link(S, ed, inf); while(bfs()) { for(int i = 0; i <= T; i++) cur[i] = head[i]; ans -= dfs(bg, inf); } printf("%d\n", ans); return 0; }