题意：给一个n*m的矩形，往每个格子填0-k的数字，使得对第i行和为row[i]，第i列和为col[i]，问是否存在方案，方案是否唯一，如果方案唯一则输出具体方案。
思路：首先根据问题提取对象，行、列、格子、数，只有数可以连接其它的对象。从源点向第i行连一条容量为row[i]的有向边，从第i行向第i行的每个格子连一条容量为k的有向边，从每个格子向第i列（i为格子的列号）连一条容量为k的有向边，然后从第i列向汇点连一条容量为col[i]的边。这样建图以后，发现每个格子唯一连接一个行和一个列，也就是入度和出度均为1，那么格子其实就没有存在的必要了，直接从每行向每列连一条容量为k的有向边。如果最大流=Σrow[i]=Σcol[i]，则表明有解。对于有多解的情况，只需判断残量网络是否存在有向环，因为在环上增减流量能得到不同的解而最大流不变。
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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 | /* ******************************************************************************** */#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; } bool FC(int fa, int rt) { vis[rt] = true; int sz = G[rt].size(); for (int i = 0; i < sz; i ++) { Edge &e = edges[G[rt][i]]; if (e.to != fa && e.cap) { if (vis[e.to]) return true; if (FC(rt, e.to)) return true; } } vis[rt] = false; return false; } bool get(int n, int m) { memset(vis, 0, sizeof(vis)); for (int i = 1; i <= n; i ++) { if (FC(-1, i)) return true; } return false; } void out(int n, int m) { int now = n * 2 + m * 2 + 1; for (int i = 0; i < n; i ++) { for (int j = 0; j < m; j ++) { printf("%d%c", edges[now].cap, j == m - 1? '
' : ' '); now += 2; } } }};Dinic solver;const int maxn = 407;int row[maxn], col[maxn];int main() {#ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin);#endif // ONLINE_JUDGE int n, m, k; while (cin >> n >> m >> k) { solver.clear(); int total = 0; for (int i = 1; i <= n; i ++) { int x; RI(x); solver.add(0, i, x); total += x; } for (int i = 1; i <= m; i ++) { int x; RI(x); solver.add(n + i, n + m + 1, x); } for (int i = 1; i <= n; i ++) { for (int j = 1; j <= m; j ++) { solver.add(i, n + j, k); } } int flow = solver.solve(0, n + m + 1); if (flow != total) puts("Impossible"); else { if (solver.get(n, m)) puts("Not Unique"); else { puts("Unique"); solver.out(n, m); } } } return 0; //} // // // ///* ******************************************************************************** */ |