POJ 1698 Alice's Chance（最大流+拆点）

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POJ 1698 Alice's Chance

题目链接

题意：拍n部电影，每部电影要在前w星期完成，并且一周只有一些天是可以拍的，每部电影有个需要的总时间，问是否能拍完电影

思路：源点向每部电影连边，容量为d，然后每部电影对应能拍的那天连边，由于每天容量限制是1，所以进行拆点，然后连向汇点即可

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;}

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