hdu2413 二分+二分匹配

题意:
      地球和外星球大战,地球有n个飞船,外星球有m个飞船,每个飞船有自己的其实战舰和战舰增长率,星球于星球之间有距离,问你最少多少年地球可以打败外星球,每个星球最多只能和一个星球对战...

思路:
      题意的最后一句话告诉我们这个题目满足二分图,我们可以二分枚举多少年打败,每次都重新建图,对于H[i] , 和 A[j],如果当前二分:  

(mid - map[i][j]) * H[i].p + H[i].st >= A[i].p * mid + A[i].st ,(地球攻击外星球,所以跑路的时间是地球付出) 就建边i,j,然后二分匹配,如果得到的匹陪数是外星战舰数,那么满足,则 up = mid - 1.........

#include<stdio.h>
#include<string.h>

#define N_node 250 + 100
#define N_edge 250 * 250 + 1000
#define INF 2000000000


typedef struct
{
   int to ,next;
}STAR;

typedef struct
{
   __int64 st ,p;
}NODE;

NODE H[N_node] ,A[N_node];
STAR E[N_edge];
__int64 map[N_node][N_node];
int mk_gx[N_node] ,mk_dfs[N_node];
int list[N_node] ,tot;

void add(int a ,int b)
{
   E[++tot].to = b;
   E[tot].next = list[a];
   list[a] = tot;
}

int DFS_XYL(int s)
{
   for(int k = list[s] ;k ;k = E[k].next)
   {
      int to = E[k].to;
      if(mk_dfs[to]) continue;
      mk_dfs[to] = 1;
      if(mk_gx[to] == -1 || DFS_XYL(mk_gx[to]))
      {
         mk_gx[to] = s;
         return 1;
      }
   }
   return 0;
}

void Buid(__int64 mid ,int n ,int m)
{
   memset(list ,0 ,sizeof(list));
   tot = 1;
   for(int i = 1 ;i <= n ;i ++)
   for(int j = 1 ;j <= m ;j ++)
   {
      __int64 h = (mid - map[i][j]) * H[i].p + H[i].st;
      __int64 a = mid * A[j].p + A[j].st;
      if(h >= a)
      add(i ,j);
   }
}

bool OK(int n ,int m)
{
   int sum = 0;
   memset(mk_gx ,255 ,sizeof(mk_gx));
   for(int i = 1 ;i <= n ;i ++)
   {
      memset(mk_dfs ,0 ,sizeof(mk_dfs));
      sum += DFS_XYL(i);
   }
   return sum == m;
}
      
   
   
int main ()
{
   int i ,j ,n ,m;
   while(~scanf("%d %d" ,&n ,&m) && n + m)
   {
      for(i = 1 ;i <= n ;i ++)
      scanf("%I64d %I64d" ,&H[i].st ,&H[i].p);
      for(i = 1 ;i <= m ;i ++)
      scanf("%I64d %I64d" ,&A[i].st ,&A[i].p);
      for(i = 1 ;i <= n ;i ++)
      for(j = 1 ;j <= m ;j ++)
      scanf("%I64d" ,&map[i][j]);
      if(n < m)
      {
         printf("IMPOSSIBLE
");
         continue;
      }
      __int64 low ,mid ,up;
      low = 0;
      up = INF;
      int ans = -1;
      while(low <= up)
      {
         mid = (low + up) >> 1;
         Buid(mid ,n ,m);
         if(OK(n ,m))
         {
            up = mid - 1;
            ans = mid;
         }
         else
         low = mid + 1;
      }
      if(ans == -1)
      printf("IMPOSSIBLE
");
      else
      printf("%d
" ,ans);
   }
   return 0;
}