题意：给定每个人所喜欢的食物和饮料种类以及每种食物和饮料的数量，一个人需要一种食物和一种饮料（数量为1即可），问最多满足多少人的需要
思路：由于食物和饮料对于人来说需要同时满足，它们是“与”的关系，所以建模时需要放在不同的层，另外如果把人放在根，食物和饮料依次放后面，则每个人会扩展出f*d个节点出来，边数有f*d条，而如果把人放中间，类似于“双向广搜”的原理，层数减半，边数大大减少。具体来说，从源点向每种食物连边，容量为其数量，如果某个人喜欢某种食物，则从食物向人连边，容量为1，为了限制人只能选择一个食物和饮料，需要人为地加n个新节点与每个人一一对应，从人向其所对应的新节点连一条容量为1的边，然后向喜欢的饮料各连一条容量为1的边，最后连回汇点，容量为饮料数量。图如下：
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 | /* ******************************************************************************** */#include <iostream> //#include <cstdio> //#include <cmath> //#include <cstdlib> //#include <cstring> //#include <vector> //#include <ctime> //#include <deque> //#include <queue> //#include <algorithm> //#include <map> //using namespace std; // //#define pb push_back //#define mp make_pair //#define X first //#define Y second //#define all(a) (a).begin(), (a).end() //#define foreach(a, i) for (typeof(a.begin()) i = a.begin(); i != a.end(); ++ i) //#define fill(a, x) memset(a, x, sizeof(a)) // //void RI(vector<int>&a,int n){a.resize(n);for(int i=0;i<n;i++)scanf("%d",&a[i]);} //void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R> //void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?1:-1; //while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T> //void print(const T t){cout<<t<<endl;}template<typename F,typename...R> //void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T> //void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;} // //typedef pair<int, int> pii; //typedef long long ll; //typedef unsigned long long ull; // //template<typename T>bool umax(T&a, const T&b){return b>a?false:(a=b,true);} //template<typename T>bool umin(T&a, const T&b){return b<a?false:(a=b,true);} //template<typename T> //void V2A(T a[],const vector<T>&b){for(int i=0;i<b.size();i++)a[i]=b[i];} //template<typename T> //void A2V(vector<T>&a,const T b[]){for(int i=0;i<a.size();i++)a[i]=b[i];} // ///* -------------------------------------------------------------------------------- */struct Dinic {private: const static int maxn = 800 + 7; struct Edge { int from, to, cap; Edge(int u, int v, int w): from(u), to(v), cap(w) {} }; int s, t; vector<Edge> edges; vector<int> G[maxn]; bool vis[maxn]; int d[maxn], cur[maxn]; bool bfs() { memset(vis, 0, sizeof(vis)); queue<int> Q; Q.push(s); d[s] = 0; vis[s] = true; while (!Q.empty()) { int x = Q.front(); Q.pop(); for (int i = 0; i < G[x].size(); i ++) { Edge &e = edges[G[x][i]]; if (!vis[e.to] && e.cap) { vis[e.to] = true; d[e.to] = d[x] + 1; Q.push(e.to); } } } return vis[t]; } int dfs(int x, int a) { if (x == t || a == 0) return a; int flow = 0, f; for (int &i = cur[x]; i < G[x].size(); i ++) { Edge &e = edges[G[x][i]]; if (d[x] + 1 == d[e.to] && (f = dfs(e.to, min(a, e.cap))) > 0) { e.cap -= f; edges[G[x][i] ^ 1].cap += f; flow += f; a -= f; if (a == 0) break; } } return flow; }public: void clear() { for (int i = 0; i < maxn; i ++) G[i].clear(); edges.clear(); memset(d, 0, sizeof(d)); } void add(int from, int to, int cap) { edges.push_back(Edge(from, to, cap)); edges.push_back(Edge(to, from, 0)); int m = edges.size(); G[from].push_back(m - 2); G[to].push_back(m - 1); } int solve(int s, int t) { this->s = s; this->t = t; int flow = 0; while (bfs()) { memset(cur, 0, sizeof(cur)); flow += dfs(s, 1e9); } return flow; }};Dinic solver;const int maxn = 207;int cf[maxn], cd[maxn];bool likef[maxn][maxn], liked[maxn][maxn];int main() {#ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin);#endif // ONLINE_JUDGE int n, f, d; while (cin >> n >> f >> d) { RI(cf + 1, cf + 1 + f); RI(cd + 1, cd + 1 + d); for (int i = 1; i <= n; i ++) { char s[234]; scanf("%s", s); for (int j = 0; j < f; j ++) { likef[i][j + 1] = s[j] == 'Y'; } } for (int i = 1; i <= n; i ++) { char s[234]; scanf("%s", s); for (int j = 0; j < d; j ++) { liked[i][j + 1] = s[j] == 'Y'; } } solver.clear(); for (int i = 1; i <= f; i ++) { solver.add(0, i, cf[i]); } for (int i = 1; i <= n; i ++) { for (int j = 1; j <= f; j ++) { if (likef[i][j]) solver.add(j, f + i, 1); } } for (int i = 1; i <= n; i ++) { solver.add(f + i, f + n + i, 1); } for (int i = 1; i <= n; i ++) { for (int j = 1; j <= d; j ++) { if (liked[i][j]) solver.add(f + n + i, f + n + n + j, 1); } } for (int i = 1; i <= d; i ++) { solver.add(f + n + n + i, f + n + n + d + 1, cd[i]); } cout << solver.solve(0, f + n + n + d + 1) << endl; } return 0; //} // // // ///* ******************************************************************************** */ |