本题是分层图最大流问题,相当于按时间拆点,每个当前点向下一点的下一时间层连点,每一层有n+1个点
#include<bits/stdc++.h> using namespace std; #define lowbit(x) ((x)&(-x)) typedef long long LL; const int maxm = 5e4+5; const int INF = 0x3f3f3f3f; struct edge{ int u, v, cap, flow, nex; } edges[maxm]; int head[maxm], cur[maxm], cnt, level[6005]; void init() { memset(head, -1, sizeof(head)); } void add(int u, int v, int cap) { edges[cnt] = edge{u, v, cap, 0, head[u]}; head[u] = cnt++; } void addedge(int u, int v, int cap) { add(u, v, cap), add(v, u, 0); } void bfs(int s) { memset(level, -1, sizeof(level)); queue<int> q; level[s] = 0; q.push(s); while(!q.empty()) { int u = q.front(); q.pop(); for(int i = head[u]; i != -1; i = edges[i].nex) { edge& now = edges[i]; if(now.cap > now.flow && level[now.v] < 0) { level[now.v] = level[u] + 1; q.push(now.v); } } } } int dfs(int u, int t, int f) { if(u == t) return f; for(int& i = cur[u]; i != -1; i = edges[i].nex) { edge& now = edges[i]; if(now.cap > now.flow && level[u] < level[now.v]) { int d = dfs(now.v, t, min(f, now.cap - now.flow)); if(d > 0) { now.flow += d; edges[i^1].flow -= d; return d; } } } return 0; } int dinic(int s, int t) { int maxflow = 0; for(;;) { bfs(s); if(level[t] < 0) break; memcpy(cur, head, sizeof(head)); int f; while((f = dfs(s, t, INF)) > 0) maxflow += f; } return maxflow; } int H[25], r[25], station[25][25]; void run_case() { init(); int n, m, k; cin >> n >> m >> k; for(int i = 1; i <= m; ++i) { cin >> H[i] >> r[i]; for(int j = 0; j < r[i]; ++j) { cin >> station[i][j]; if(station[i][j] == -1) station[i][j] = n+1; } } int s = 0, t = n+1, maxflow = 0; for(int times = 1; times <= 500; ++times) { for(int i = 0; i <= n; ++i) addedge(i+(times-1)*(n+2), i+(times)*(n+2), INF); addedge(t+times*(n+2), t+(times-1)*(n+2), INF); for(int i = 1; i <= m; ++i) addedge((times-1)*(n+2)+station[i][(times-1)%r[i]], times*(n+2)+station[i][times%r[i]], H[i]); maxflow += dinic(s, t); if(maxflow >= k) { cout << times; return; } } cout << 0; } int main() { ios::sync_with_stdio(false), cin.tie(0); run_case(); cout.flush(); return 0; }