题意：  给你一个N ，M   构造一个N*M的矩阵，矩阵中每个元素为0-K；

给你每行的和与每列的和。

如果解法唯一 ，输出解法

如果解法不唯一，输出一句话，

如果没有解法，输出一句话。

题解：   经典建图

             s ---> 每个行节点，流量为行和

             每个列节点----〉t，流量为列和

            每行每列单独连接，流量为K 

代码：

#include<stdio.h>
#include<queue>
#include<iostream>
#define INF 1000000000
#define N 1005
using namespace std;
int list[N], listt[N], deep[N], tot, n, m, k, flagg, mark_dfs[N];
struct Node 
{
   int date, value, next;      
}cun[2000005];
struct Edge
{
   int x, t;       
}old, xin;
void add(int a, int b, int c)
{
   cun[++tot].date = b;
   cun[tot].value = c;
   cun[tot].next = list[a];
   list[a] = tot;
   
   cun[++tot].date = a;
   cun[tot].value = 0;
   cun[tot].next = list[b];
   list[b] = tot;     
}
int bfs(int s, int t, int n)
{
   queue<Edge> p;
   xin.x = s;
   xin.t = 0;
   p.push(xin);
   memset(deep,255,sizeof(deep));
   deep[s] = 0;
   while(!p.empty())
   {
      old = p.front();
      p.pop();
      for(int i = list[old.x]; i; i = cun[i].next)
      {
         int date = cun[i].date;
         int value = cun[i].value;
         xin.x = date;
         xin.t = old.t + 1;
         if(value == 0 || deep[date] != -1)  continue;
         deep[date] = xin.t;
         p.push(xin);              
      }                        
   } 
  for(int i = 0; i <= n; i++)
  {
    listt[i] = list[i];        
  }  
  return deep[t] != -1;
}
int minn( int a, int b)
{
   if(a < b)  return a;
   return b;    
}
int dfs(int s, int t, int min)
{
  if(s == t) return min;
  int neww = 0;
  for(int i = listt[s]; i; i = cun[i].next)
  {
    listt[s] = i;
    int date = cun[i].date;
    int value = cun[i].value;
    if(value == 0 || deep[date] != deep[s] + 1)  { continue; }  
    int m = dfs(date, t, minn(value, min - neww));
    neww += m;
    cun[i].value -= m;
    cun[i^1].value += m;
    if(neww == min)   break;      
  }    
  if(neww == 0)   deep[s] = 0;
  return neww;
}
int  dinic( int s, int t, int n)
{
   int num = 0;
   while(bfs(s, t, n))
   {
     num += dfs(s, t, INF);             
   }     
   return num;
}
int dfs_1(int s, int p)
{
   for(int i = list[s]; i; i = cun[i].next)
   {
      int date = cun[i].date;
      if(i == (p^1))  continue;
      if(cun[i].value)
      {
       if(mark_dfs[date])    {return 1;}
       mark_dfs[date] = 1;        
       if(dfs_1(date, i))  return 1;
       mark_dfs[date] = 0;
      }
   } 
   return 0;  
}
void build()
{
      int a, s = n + m + 10, t = s + 1, sum1 = 0, sum2 = 0;
      memset(list,0,sizeof(list));
      memset(mark_dfs,0,sizeof(mark_dfs));
      tot = 1;
      flagg = 0;
      for(int i = 1; i <= n; i++ )
      {
        scanf("%d", &a);
        sum1 += a;
        add(s, i, a);        
      }              
      for(int i = 1;i <= m; i++ )
      {
        scanf("%d", &a);
        sum2 += a;
        add(i + n, t, a);        
      }           
      if(sum1 != sum2)  {printf("Impossible\n");  return;}
      int kk = tot;
      for(int i = 1; i <= n; i++ )
      for(int j = 1; j <= m; j++ )
      {
        add(i, j + n, k);        
      }  
      int mm = dinic ( s, t, t + 10 ), flag = 0;
      if(mm != sum1) {printf("Impossible\n");  return;}
      for(int i = 1; i <= n; i++ )
      {
        if( dfs_1(i, -1) )
        {
           flagg = 1;
           break;    
        }
      } 
      kk++;
      if(flagg)   printf("Not Unique\n");
      else
      {
         printf("Unique\n");
         for(int i = 1; i <= n; i++)
           {
            for(int j = 1; j <= m; j++)
            {
             if(j == 1)  printf("%d",k - cun[kk].value);
             else printf(" %d",k - cun[kk].value);
             kk += 2;        
            } 
            printf("\n");
           }
      }
      
}
int main()
{
   while(scanf("%d%d%d",&n, &m, &k)!=EOF)
   {
      build();
   }    
}

HDU 4888 神奇最大流行进列出构造矩阵