POJ 1698 Alice‘s Chance
题意:拍n部电影。每部电影要在前w星期完毕,而且一周仅仅有一些天是能够拍的,每部电影有个须要的总时间,问能否拍完电影
思路:源点向每部电影连边,容量为d,然后每部电影相应能拍的那天连边,因为每天容量限制是1。所以进行拆点,然后连向汇点就可以
代码:
#include <cstdio> #include <cstring> #include <queue> #include <algorithm> using namespace std; const int MAXNODE = 1005; const int MAXEDGE = 100005; typedef int Type; const Type INF = 0x3f3f3f3f; struct Edge { int u, v; Type cap, flow; Edge() {} Edge(int u, int v, Type cap, Type flow) { this->u = u; this->v = v; this->cap = cap; this->flow = flow; } }; struct Dinic { int n, m, s, t; Edge edges[MAXEDGE]; int first[MAXNODE]; int next[MAXEDGE]; bool vis[MAXNODE]; Type d[MAXNODE]; int cur[MAXNODE]; vector<int> cut; void init(int n) { this->n = n; memset(first, -1, sizeof(first)); m = 0; } void add_Edge(int u, int v, Type cap) { edges[m] = Edge(u, v, cap, 0); next[m] = first[u]; first[u] = m++; edges[m] = Edge(v, u, 0, 0); next[m] = first[v]; first[v] = m++; } bool bfs() { memset(vis, false, sizeof(vis)); queue<int> Q; Q.push(s); d[s] = 0; vis[s] = true; while (!Q.empty()) { int u = Q.front(); Q.pop(); for (int i = first[u]; i != -1; i = next[i]) { Edge& e = edges[i]; if (!vis[e.v] && e.cap > e.flow) { vis[e.v] = true; d[e.v] = d[u] + 1; Q.push(e.v); } } } return vis[t]; } Type dfs(int u, Type a) { if (u == t || a == 0) return a; Type flow = 0, f; for (int &i = cur[u]; i != -1; i = next[i]) { Edge& e = edges[i]; if (d[u] + 1 == d[e.v] && (f = dfs(e.v, min(a, e.cap - e.flow))) > 0) { e.flow += f; edges[i^1].flow -= f; flow += f; a -= f; if (a == 0) break; } } return flow; } Type Maxflow(int s, int t) { this->s = s; this->t = t; Type flow = 0; while (bfs()) { for (int i = 0; i < n; i++) cur[i] = first[i]; flow += dfs(s, INF); } return flow; } void MinCut() { cut.clear(); for (int i = 0; i < m; i += 2) { if (vis[edges[i].u] && !vis[edges[i].v]) cut.push_back(i); } } } gao; int t, n, day[10], d, w, vis[400]; int main() { scanf("%d", &t); while (t--) { memset(vis, 0, sizeof(vis)); gao.init(1000); scanf("%d", &n); int sum = 0; for (int i = 1; i <= n; i++) { for (int j = 1; j <= 7; j++) scanf("%d", &day[j]); scanf("%d%d", &d, &w); sum += d; gao.add_Edge(0, i, d); for (int j = 0; j < w; j++) { for (int k = 1; k <= 7; k++) { if (day[k]) { gao.add_Edge(i, 20 + j * 7 + k, INF); vis[20 + j * 7 + k] = 1; } } } } for (int i = 21; i <= 370; i++) { if (!vis[i]) continue; gao.add_Edge(i, i + 350, 1); gao.add_Edge(i + 350, 1000, INF); } printf("%s\n", gao.Maxflow(0, 1000) == sum ? "Yes" : "No"); } return 0; }
POJ 1698 Alice's Chance(最大流+拆点)
时间: 2024-10-12 06:01:44