hdu 3046(最小割）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3046
思路：最小割的入门题，设源点为0，汇点为n*m+1,源点与点为2的连一天容量为inf的边，汇点与点为1的连容量为inf的边，每个相邻网格连容量为1的边。
View Code
 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 using namespace std;
 5 #define MAXN 44444
 6 #define MAXM 555555
 7 #define inf 1<<30
 8 struct Edge{
 9     int v,cap,next;
10 }edge[MAXM];
11 
12 int head[MAXN];
13 int pre[MAXN];
14 int cur[MAXN];
15 int level[MAXN];
16 int gap[MAXN];
17 int NV,NE,n,m,vs,vt;
18 int dir[4][2]={{0,-1},{0,1},{-1,0},{1,0}};//up,down,left,right
19 
20 void Insert(int u,int v,int cap,int cc=0){
21     edge[NE].v=v;edge[NE].cap=cap;
22     edge[NE].next=head[u];head[u]=NE++;
23 
24     edge[NE].v=u;edge[NE].cap=cc;
25     edge[NE].next=head[v];head[v]=NE++;
26 }
27 
28 
29 int SAP(int vs,int vt){
30     memset(pre,-1,sizeof(pre));
31     memset(level,0,sizeof(level));
32     memset(gap,0,sizeof(gap));
33     for(int i=0;i<=n*m+1;i++)cur[i]=head[i];
34     int u=pre[vs]=vs,maxflow=0,aug=-1;
35     gap[0]=NV;
36     while(level[vs]<NV){
37 loop:
38         for(int &i=cur[u];i!=-1;i=edge[i].next){
39             int v=edge[i].v;
40             if(edge[i].cap&&level[u]==level[v]+1){
41                 aug==-1?aug=edge[i].cap:(aug=min(aug,edge[i].cap));
42                 pre[v]=u;
43                 u=v;
44                 if(v==vt){
45                     maxflow+=aug;
46                     for(u=pre[u];v!=vs;v=u,u=pre[u]){
47                         edge[cur[u]].cap-=aug;
48                         edge[cur[u]^1].cap+=aug;
49                     }
50                     aug=-1;
51                 }
52                 goto loop;
53             }
54         }
55         int minlevel=NV;
56         for(int i=head[u];i!=-1;i=edge[i].next){
57             int v=edge[i].v;
58             if(edge[i].cap&&minlevel>level[v]){
59                 cur[u]=i;
60                 minlevel=level[v];
61             }
62         }
63         gap[level[u]]--;
64         if(gap[level[u]]==0)break;
65         level[u]=minlevel+1;
66         gap[level[u]]++;
67         u=pre[u];
68     }
69     return maxflow;
70 }
71 
72 
73 int main(){
74     int _case=1,tmp;
75     while(~scanf("%d%d",&n,&m)){
76         NE=0,NV=n*m+2,vs=0,vt=n*m+1;
77         memset(head,-1,sizeof(head));
78         for(int i=1;i<=n;i++){
79             for(int j=1;j<=m;j++){
80                 scanf("%d",&tmp);
81                 if(tmp==1){Insert((i-1)*m+j,vt,inf);}
82                 else if(tmp==2){Insert(vs,(i-1)*m+j,inf);}
83                 for(int k=0;k<4;k++){
84                     int x=i+dir[k][0];
85                     int y=j+dir[k][1];
86                     if(x>=1&&x<=n&&y>=1&&y<=m){Insert((i-1)*m+j,(x-1)*m+y,1);}
87                 }
88             }
89         }
90         printf("Case %d:\n",_case++);
91         printf("%d\n",SAP(vs,vt));
92     }
93     return 0;
94 }