题目链接:HDU-4292 Food
题意
有$N$个人、$F$种食物、$D$种饮料,每种食物和饮料有一定数量,每个人对每种食物和饮料有数量为1的需求或无需求,问这$F$种食物和$D$种饮料最多能满足多少个人的需求。
思路
一个人拆成两个结点,分为左部点和右部点,左部点向右部点连容量为1的边,表示一个人对答案的贡献最多为1;
源点向代表食物的结点连边,容量为对应的食物的数量,食物向对这种食物有需求的人的左部点连边,容量为1,表示需求为1;
人的右部点向有需求的饮料连边,容量为1,表示对这种饮料需求为1,饮料向汇点连边,容量为对应的饮料的数量,然后跑最大流即可。
代码实现
#include <iostream> #include <cstdio> #include <cstring> #include <queue> using std::queue; const int INF = 0x3f3f3f3f, N = 1000, M = 200000; int head[N], d[N]; int s, t, tot, maxflow; struct Edge { int to, cap, nex; } edge[M]; queue<int> q; void add(int x, int y, int z) { edge[++tot].to = y, edge[tot].cap = z, edge[tot].nex = head[x], head[x] = tot; edge[++tot].to = x, edge[tot].cap = 0, edge[tot].nex = head[y], head[y] = tot; } bool bfs() { memset(d, 0, sizeof(d)); while (q.size()) q.pop(); q.push(s); d[s] = 1; while (q.size()) { int x = q.front(); q.pop(); for (int i = head[x]; i; i = edge[i].nex) { int v = edge[i].to; if (edge[i].cap && !d[v]) { q.push(v); d[v] = d[x] + 1; if (v == t) return true; } } } return false; } int dinic(int x, int flow) { if (x == t) return flow; int rest = flow, k; for (int i = head[x]; i && rest; i = edge[i].nex) { int v = edge[i].to; if (edge[i].cap && d[v] == d[x] + 1) { k = dinic(v, std::min(rest, edge[i].cap)); if (!k) d[v] = 0; edge[i].cap -= k; edge[i^1].cap += k; rest -= k; } } return flow - rest; } void init(int v_num) { tot = 1, maxflow = 0; s = v_num, t = s + 1; memset(head, 0, sizeof(head)); } int main() { int nn, nf, nd; while (~scanf("%d %d %d", &nn, &nf, &nd)) { init(nn * 2 + nf + nd); for (int i = 0, num; i < nf; i++) { scanf("%d", &num); add(s, i, num); } for (int i = 0, num; i < nd; i++) { scanf("%d", &num); add(i + nf, t, num); } char str[210]; for (int i = 0; i < nn; i++) { scanf(" %s", str); for (int j = 0; j < nf; j++) { if (str[j] == 'Y') add(j, i + nf + nd, 1); } add(nf + nd + i, nf + nd + nn + i, 1); } for (int i = 0; i < nn; i++) { scanf(" %s", str); for (int j = 0; j < nd; j++) { if (str[j] == 'Y') add(nn + nf + nd + i, nf + j, 1); } } while (bfs()) maxflow += dinic(s, INF); printf("%d\n", maxflow); } return 0; }