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 }