hdu2413 二分+二分匹配

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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 2000000000typedef 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\n");         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\n");      else      printf("%d\n" ,ans);   }   return 0;}

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