HDU 3360 Minimum Point coverage

Test instructions: Give you a picture, there are two elements of treasure and security. Each treasure needs to be placed at the same time in some locations around the security (if those locations have treasures, you can replace the treasure to security) ask you at least need to place more security, you can make all the treasures meet the requirements.

Test instructions's a little hard to understand.

Link: Point Me

Direct construction of the non-directed graph, less judgment of the parity, and finally a 2 can be

1#include <cstdio>2#include <iostream>3#include <algorithm>4#include <cstring>5#include <cmath>6#include <queue>7#include <map>8 using namespacestd;9 #defineMOD 1000000007Ten Const intinf=0x3f3f3f3f; One Const Doubleeps=1e-5; AtypedefLong Longll; - #defineCL (a) memset (A,0,sizeof (a)) - #defineTS printf ("*****\n"); the Const intMAXN =5010;//maximum number of points - Const intMAXM =50010;//maximum number of edges - intA[MAXN][MAXN]; - intB[MAXN][MAXN]; + intN,m,tt; - /* + * Hungarian algorithm adjacency table Form A * Initialize with init () before use, assign value to UN at * Add edge using function Addedge (u,v) - * - */ - structEdge - { -     intTo,next; in }EDGE[MAXM]; - intHead[maxn],tot; to voidInit () + { -tot =0; thememset (head,-1,sizeof(head)); * } $ voidAddedge (intUintv)Panax Notoginseng { -Edge[tot].to = v; Edge[tot].next =Head[u]; theHead[u] = tot++; + } A intLINKER[MAXN]; the BOOLUSED[MAXN]; + intUN; - BOOLDfsintu) $ { $      for(inti = Head[u]; I! =-1; i =edge[i].next) -     { -         intv =edge[i].to; the         if(!Used[v]) -         {WuyiUSED[V] =true; the             if(Linker[v] = =-1||DFS (Linker[v])) -             { WuLINKER[V] =u; -                 return true; About             } $         } -     } -     return false; - } A intHungary () + { the     intres =0; -memset (linker,-1,sizeof(linker)); $      for(intU =0; U < un;u++)//number of points 0~un-1 the     { thememset (Used,false,sizeof(used)); the         if(Dfs (U)) res++; the     } -     returnRes; in } the intdir[][2] = {{-1,-2},{-2,-1},{-2,1},{-1,2},{1,2}, the{2,1},{2,-1},{1,-2},{-1,0},{0,1},{1,0},{0,-1}}; About intMain () the { the     inti,j,k; the #ifndef Online_judge +Freopen ("1.in","R", stdin); -     #endif the     intCa=1;Bayi      while(SCANF ("%d%d", &n,&m)! =EOF) the     { the         if(n==0&&m==0) Break; -UN =0; -          for(i =0; I < n;i++) the              for(j =0; J < m;j++) the             { theB[I][J] = un++; the             } -          for(i=0; i<n;i++) the         { the              for(j=0; j<m;j++) the             {94scanf"%d",&a[i][j]); the             } the         } the init ();98          for(i=0; i<n;i++) About         { -              for(j=0; j<m;j++)101             {102                 if(a[i][j]!=-1)103                 {104                      for(k=0;k< A; k++) the                     {106                         if(a[i][j]& (1&LT;&LT;K))//this point needs to be placed on guard107                         {108                             intnx=i+dir[k][0];109                             intny=j+dir[k][1]; the                             if(nx>=0&&nx<n&&ny>=0&&ny<m&&a[nx][ny]!=-1)111                             { the Addedge (B[i][j],b[nx][ny]);113 Addedge (B[nx][ny],b[i][j]); the                             } the                         } the                     }117                 }118             }119         } -printf"%d.%d\n", Ca++,hungary ()/2);121     }122}

HDU 3360 Minimum Point coverage