题意：给一个N行M列的数字矩阵的行和以及列和，每个元素的大小不超过K，问这样的矩阵是否存在，是否唯一，唯一则求出各个元素N(1 ≤ N ≤ 400) , M(1 ≤ M ≤ 400), K(1 ≤ K ≤ 40)。
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4888
——>>建图：
1）超级源S = 0，超级汇T = N + M + 1；
2）S到每个行和各连一条边，容量为该行行和；
3）每个行和到每个列和各连一条边，容量为K；
4）每个列和到 T 各连一条边，容量为该列列和。
一个行到所有列连边，为的是让该行分流多少给各个列，正是该行某列元素的大小。。
所以，如果 S 到 T 的最大流 == 所有元素的和，则说明有解。。
残量网络中的行列结点之间如果有长度 > 2 的环（自环长度为2，但无法调整流量），则说明这个环中的流量可以调整，使得达到最大流时该环上的流量不唯一，即矩阵不唯一。。
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>using std::min;
using std::queue;const int MAXN = 400 * 2 + 10;
const int MAXM = 400 * 400 + 2 * MAXN;
const int INF = 0x3f3f3f3f;struct EDGE
{int to;int cap;int flow;int nxt;
} edge[MAXM << 1];int N, M, K;
int sum;
int S, T;
int hed[MAXN], ecnt;
int cur[MAXN], h[MAXN];
bool impossible, bUnique;void Init()
{impossible = false;bUnique = true;ecnt = 0;memset(hed, -1, sizeof(hed));
}void AddEdge(int u, int v, int cap)
{edge[ecnt].to = v;edge[ecnt].cap = cap;edge[ecnt].flow = 0;edge[ecnt].nxt = hed[u];hed[u] = ecnt++;edge[ecnt].to = u;edge[ecnt].cap = 0;edge[ecnt].flow = 0;edge[ecnt].nxt = hed[v];hed[v] = ecnt++;
}bool Bfs()
{memset(h, -1, sizeof(h));queue<int> qu;qu.push(S);h[S] = 0;while (!qu.empty()){int u = qu.front();qu.pop();for (int e = hed[u]; e != -1; e = edge[e].nxt){int v = edge[e].to;if (h[v] == -1 && edge[e].cap > edge[e].flow){h[v] = h[u] + 1;qu.push(v);}}}return h[T] != -1;
}int Dfs(int u, int cap)
{if (u == T || cap == 0) return cap;int flow = 0, subFlow;for (int e = cur[u]; e != -1; e = edge[e].nxt){cur[u] = e;int v = edge[e].to;if (h[v] == h[u] + 1 && (subFlow = Dfs(v, min(cap, edge[e].cap - edge[e].flow))) > 0){flow += subFlow;edge[e].flow += subFlow;edge[e ^ 1].flow -= subFlow;cap -= subFlow;if (cap == 0) break;}}return flow;
}int Dinic()
{int maxFlow = 0;while (Bfs()){memcpy(cur, hed, sizeof(hed));maxFlow += Dfs(S, INF);}return maxFlow;
}void Read()
{int r, c;int rsum = 0, csum = 0;S = 0;T = N + M + 1;for (int i = 1; i <= N; ++i){scanf("%d", &r);rsum += r;AddEdge(S, i, r);}for (int i = 1; i <= M; ++i){scanf("%d", &c);csum += c;AddEdge(i + N, T, c);}if (rsum != csum){impossible = true;return;}sum = rsum;for (int i = 1; i <= N; ++i){for (int j = M; j >= 1; --j){AddEdge(i, j + N, K);}}
}void CheckPossible()
{if (impossible) return;if (Dinic() != sum){impossible = true;}
}bool vis[MAXN];
bool CheckCircle(int x, int f)
{vis[x] = true;for (int e = hed[x]; e != -1; e = edge[e].nxt){if (edge[e].cap > edge[e].flow){int v = edge[e].to;if (v == f) continue;if (vis[v]) return true;else{if (CheckCircle(v, x)) return true;}}}vis[x] = false;return false;
}void CheckUnique()
{if (impossible) return;memset(vis, 0, sizeof(vis));for (int i = 1; i <= N; ++i){if (CheckCircle(i, -1)){bUnique = false;return;}}
}void Output()
{if (impossible){puts("Impossible");}else if (!bUnique){puts("Not Unique");}else{puts("Unique");for (int i = 1; i <= N; ++i){for (int e = hed[i], j = 1; e != -1 && j <= M; e = edge[e].nxt, ++j){printf("%d", edge[e].flow);if (j < M){printf(" ");}}puts("");}}
}int main()
{while (scanf("%d%d%d", &N, &M, &K) == 3){Init();Read();CheckPossible();CheckUnique();Output();}return 0;
}