UVA 10779 - Collectors Problem（网络流）

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UVA 10779 - Collectors Problem

题目链接

题意：每个人有一种贴图，现在第0个人要去和别人交换贴图，来保证自己的贴图尽量多，只有别人没有该种贴图，并且自己有2张以上另一种贴图才会换，问最多有几张贴图

思路：最大流，关键在于如何建模，把0号人和物品连边，容量为有的容量，然后其他人如果物品等于0的，连一条边从物品到这个人，表示能交换，然后如果物品大于1的，连一条边从这个人到物品，容量为物品减1（自己要留一个），然后把所有物品连到汇点，跑一次最大流即可

代码：

#include <cstdio>#include <cstring>#include <queue>#include <algorithm>using namespace std;const int MAXNODE = 1005;const int MAXEDGE = 100005;const int INF = 0x3f3f3f3f;struct Edge {int u, v, cap, flow;Edge() {}Edge(int u, int v, int cap, int 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];int d[MAXNODE];int cur[MAXNODE];void init(int n) {this->n = n;memset(first, -1, sizeof(first));m = 0;}void add_Edge(int u, int v, int 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];}int dfs(int u, int a) {if (u == t || a == 0) return a;int 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;}int Maxflow(int s, int t) {this->s = s;this->t = t;int flow = 0;while (bfs()) {for (int i = 0; i < n; i++)cur[i] = first[i];flow += dfs(s, INF);}return flow;}} gao;const int N = 30;int T, n, m, cnt[N];int main() {int cas = 0;scanf("%d", &T);while (T--) {scanf("%d%d", &n, &m);int s = 0, t = n + m;gao.init(t + 1);for (int i = 0; i < n; i++) {int tot, tmp;memset(cnt, 0, sizeof(cnt));scanf("%d", &tot);while (tot--) {scanf("%d", &tmp);cnt[tmp]++;}if (i == 0) {for (int j = 1; j <= m; j++)if (cnt[j] > 0) gao.add_Edge(i, j + n - 1, cnt[j]);} else {for (int j = 1; j <= m; j++) {if (cnt[j] > 1) gao.add_Edge(i, j + n - 1, cnt[j] - 1);if (cnt[j] == 0) gao.add_Edge(j + n - 1, i, 1);}}}for (int i = 1; i <= m; i++)gao.add_Edge(i + n - 1, t, 1);printf("Case #%d: %d\n", ++cas, gao.Maxflow(s, t));}return 0;}

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