Leapin' Lizards(hdu 2732)

Leapin' Lizards

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2080    Accepted Submission(s): 857

Problem Description

Your
platoon of wandering lizards has entered a strange room in the
labyrinth you are exploring. As you are looking around for hidden
treasures, one of the rookies steps on an innocent-looking stone and the
room's floor suddenly disappears! Each lizard in your platoon is left
standing on a fragile-looking pillar, and a fire begins to rage below...
Leave no lizard behind! Get as many lizards as possible out of the
room, and report the number of casualties.
The pillars in the room
are aligned as a grid, with each pillar one unit away from the pillars
to its east, west, north and south. Pillars at the edge of the grid are
one unit away from the edge of the room (safety). Not all pillars
necessarily have a lizard. A lizard is able to leap onto any unoccupied
pillar that is within d units of his current one. A lizard standing on a
pillar within leaping distance of the edge of the room may always leap
to safety... but there's a catch: each pillar becomes weakened after
each jump, and will soon collapse and no longer be usable by other
lizards. Leaping onto a pillar does not cause it to weaken or collapse;
only leaping off of it causes it to weaken and eventually collapse. Only
one lizard may be on a pillar at any given time.

 

Input

The
input file will begin with a line containing a single integer
representing the number of test cases, which is at most 25. Each test
case will begin with a line containing a single positive integer n
representing the number of rows in the map, followed by a single
non-negative integer d representing the maximum leaping distance for the
lizards. Two maps will follow, each as a map of characters with one row
per line. The first map will contain a digit (0-3) in each position
representing the number of jumps the pillar in that position will
sustain before collapsing (0 means there is no pillar there). The second
map will follow, with an 'L' for every position where a lizard is on
the pillar and a '.' for every empty pillar. There will never be a
lizard on a position where there is no pillar.Each input map is
guaranteed to be a rectangle of size n x m, where 1 ≤ n ≤ 20 and 1 ≤ m ≤
20. The leaping distance is
always 1 ≤ d ≤ 3.

 

Output

For
each input case, print a single line containing the number of lizards
that could not escape. The format should follow the samples provided
below.

 思路：最大流，将每个点拆分成两个，一个为入流点一个为出流点，两点间建立流量为这个点处柱子的高度。

然后建立一个总输出点0，和总汇点，还有在两个距离不超过d的点之间建立流量为无穷的边，0和每个L的输入点建立流为1的边。最后跑下dinic。用所有的L-最大流即可。

  1 #include<stdio.h>
  2 #include<algorithm>
  3 #include<stdlib.h>
  4 #include<iostream>
  5 #include<string.h>
  6 #include<vector>
  7 #include<queue>
  8 using namespace std;
  9 struct node
 10 {
 11         int to;
 12         int cap;
 13         int rev;
 14 };
 15 char str[30][30];
 16 char  str2[30][30];
 17 int id[30][30];
 18 int level[100000];
 19 vector<node>vec[100000];
 20 const int N=1e8;
 21 int iter[100000];
 22 void add(int from,int to,int cap);
 23 void bfs(int s);
 24 int dfs(int s,int t,int f);
 25 int max_flow(int s,int t);
 26 int  main(void)
 27 {
 28         int i,j,k;
 29         scanf("%d",&k);
 30         int s;
 31         int p,q;
 32         for(s=1; s<=k; s++)
 33         {
 34                 scanf("%d %d ",&p,&q);
 35                 for(i=0; i<100000; i++)
 36                         vec[i].clear();
 37                 for(i=0; i<p; i++)
 38                 {
 39                         scanf("%s",str[i]);
 40                 }
 41                 for(i=0; i<p; i++)
 42                 {
 43                         scanf("%s",str2[i]);
 44                 }
 45                 int l=strlen(str[0]);
 46                 int ans=1;
 47                 for(i=0; i<p; i++)
 48                 {
 49                         for(j=0; j<l; j++)
 50                         {
 51                                 id[i][j]=ans++;
 52                         }
 53                 }
 54                 ans-=1;
 55                 int x,y;
 56                 for(i=0; i<p; i++)
 57                 {
 58                         for(j=0; j<l; j++)
 59                         {
 60                                 if(str[i][j]>'0')
 61                                         add(id[i][j],id[i][j]+ans,str[i][j]-'0');
 62                         }
 63                 }
 64                 for(i=0; i<p; i++)
 65                 {
 66                         for(j=0; j<l; j++)
 67                         {
 68                                 for(x=0; x<p; x++)
 69                                 {
 70                                         for(y=0; y<l; y++)
 71                                         {
 72                                                 if(i!=x||j!=y)
 73                                                 {
 74                                                         if(abs(x-i)+abs(j-y)<=q)
 75                                                         {
 76                                                                 add(id[i][j]+ans,id[x][y],N);
 77                                                         }
 78                                                 }
 79                                         }
 80                                 }
 81                         }
 82                 }
 83                 for(i=0; i<p; i++)
 84                 {
 85                         for(j=0; j<l; j++)
 86                         {
 87                                 int cc=abs(i+1);
 88                                 int dd=abs(p-i);
 89                                 cc=min(cc,dd);
 90                                 int kk=abs(j+1);
 91                                 int vv=abs(l-j);
 92                                 vv=min(kk,vv);
 93                                 if((vv<=q||cc<=q))
 94                                 {
 95                                         add(id[i][j]+ans,2*ans+1,N);
 96                                 }
 97                         }
 98                 }
 99                 int fuck=0;
100                 for(i=0; i<p; i++)
101                 {
102                         for(j=0; j<l; j++)
103                         {
104                                 if(str2[i][j]=='L')
105                                 {
106                                         add(0,id[i][j],1);
107                                         fuck++;
108                                 }
109                         }
110                 }
111                 int sum=max_flow(0,2*ans+1);
112                 printf("Case #%d: ",s);
113                 if(fuck-sum==1)
114                 printf("%d lizard was left behind.\n",fuck-sum);
115                 else if(fuck-sum==0)  printf("no lizard was left behind.\n");
116                 else  printf("%d lizards were left behind.\n",fuck-sum);
117         }
118 }
119 void add(int from,int to,int cap)
120 {
121         node nn;
122         nn.to=to;
123         nn.cap=cap;
124         nn.rev=vec[to].size();
125         vec[from].push_back(nn);
126         nn.to=from;
127         nn.cap=0;
128         nn.rev=vec[from].size()-1;
129         vec[to].push_back(nn);
130 }
131 void bfs(int s)
132 {
133         queue<int>que;
134         memset(level,-1,sizeof(level));
135         level[s]=0;
136         que.push(s);
137         while(!que.empty())
138         {
139                 int v=que.front();
140                 que.pop();
141                 int i;
142                 for(i=0; i<vec[v].size(); i++)
143                 {
144                         node e=vec[v][i];
145                         if(level[e.to]==-1&&e.cap>0)
146                         {
147                                 level[e.to]=level[v]+1;
148                                 que.push(e.to);
149                         }
150                 }
151         }
152 }
153 int dfs(int s,int t,int f)
154 {
155         if(s==t)
156                 return f;
157         for(int &i=iter[s]; i<vec[s].size(); i++)
158         {
159                 node &e=vec[s][i];
160                 if(level[e.to]>level[s]&&e.cap>0)
161                 {
162                         int r=dfs(e.to,t,min(e.cap,f));
163                         if(r>0)
164                         {
165                                 e.cap-=r;
166                                 vec[e.to][e.rev].cap+=r;
167                                 return r;
168                         }
169                 }
170         }
171         return 0;
172 }
173 int max_flow(int s,int t)
174 {
175         int flow=0;
176         for(;;)
177         {
178                 bfs(s);
179                 if(level[t]<0)return flow;
180                 memset(iter,0,sizeof(iter));
181                 int f;
182                 while((f=dfs(s,t,N))>0)
183                 {
184                         flow+=f;
185                 }
186         }
187 }