hdu 3046 最小割

 

每个栅栏其实就是一条边,修一些栅栏,使得狼不能抓到羊,其实就是求一个割,使得羊全在S中,狼全在T中。

 

  1 #include <cstdio>
  2 #include <cstring>
  3 #include <vector>
  4 #include <queue>
  5 #define maxn 40010
  6 #define oo 0x3f3f3f3f
  7 #define clr(arr,n) memset(&arr,0,sizeof(arr[0])*(n+1))
  8 using namespace std;
  9 
 10 struct Edge {
 11     int u, v, f;
 12     Edge( int u, int v, int f ):u(u),v(v),f(f){}
 13 };
 14 struct Dinic {
 15     int n, src, dst;
 16     vector<Edge> edge;
 17     vector<int> g[maxn];
 18     int dep[maxn], cur[maxn];
 19     
 20     void init( int n, int src, int dst ) {
 21         this->n = n;
 22         this->src = src;
 23         this->dst = dst;
 24         for( int u=1; u<=n; u++ )
 25             g[u].clear();
 26         edge.clear();
 27     }
 28     void add_edge( int u, int v, int f ) {
 29         g[u].push_back( edge.size() );
 30         edge.push_back( Edge(u,v,f) );
 31         g[v].push_back( edge.size() );
 32         edge.push_back( Edge(v,u,0) );
 33     }
 34     bool bfs() {
 35         queue<int> qu;
 36         clr( dep, n );
 37         qu.push(src);
 38         dep[src] = 1;
 39         while( !qu.empty() ) {
 40             int u=qu.front();
 41             qu.pop();
 42             for( int t=0; t<g[u].size(); t++ ) {
 43                 Edge &e=edge[g[u][t]];
 44                 if( e.f && !dep[e.v] ) {
 45                     dep[e.v] = dep[e.u]+1;
 46                     qu.push( e.v );
 47                 }
 48             }
 49         }
 50         return dep[dst];
 51     }
 52     int dfs( int u, int a ) {
 53         if( u==dst || a==0 ) return a;
 54         int remain=a, past=0, na;
 55         for( int &t=cur[u]; t<g[u].size(); t++ ) {
 56             Edge &e = edge[g[u][t]];
 57             Edge &ve = edge[g[u][t]^1];
 58             if( e.f && dep[e.v]==dep[e.u]+1 && (na=dfs(e.v,min(e.f,remain))) ) {
 59                 remain -= na;
 60                 past += na;
 61                 e.f -= na;
 62                 ve.f += na;
 63                 if( remain==0 ) break;
 64             }
 65         }
 66         return past;
 67     }
 68     int maxflow() {
 69         int flow = 0;
 70         while(bfs()) {
 71             clr( cur, n );
 72             flow += dfs(src,oo);
 73         }
 74         return flow;
 75     }
 76 };
 77 
 78 int n, m;
 79 int idx[210][210], id_clock;
 80 int map[210][210];
 81 int dx[2] = { +1, 0 };
 82 int dy[2] = { 0, +1 };
 83 Dinic D;
 84 
 85 int main() {
 86     for( int cas=1; ; cas++ ) {
 87         if( scanf( "%d%d", &n, &m )!=2 ) return 0;
 88         id_clock = 0;
 89         for( int i=1; i<=n; i++ ) 
 90             for( int j=1; j<=m; j++ ) {
 91                 scanf( "%d", &map[i][j] );
 92                 idx[i][j] = ++id_clock;
 93             }
 94         D.init( id_clock+2, id_clock+1, id_clock+2 );
 95         for( int i=1; i<=n; i++ )
 96             for( int j=1; j<=m; j++ ) 
 97                 for( int d=0; d<2; d++ ) {
 98                     int ni = i+dx[d];
 99                     int nj = j+dy[d];
100                     if( 1<=ni&&ni<=n && 1<=nj&&nj<=m ) {
101                         int u = idx[i][j];
102                         int v = idx[ni][nj];
103                         D.add_edge( u, v, 1 );
104                         D.add_edge( v, u, 1 );
105                     }
106                 }
107         for( int i=1; i<=n; i++ )
108             for( int j=1; j<=m; j++ ) {
109                 if( map[i][j]==1 ) 
110                     D.add_edge( D.src, idx[i][j], oo );
111                 if( map[i][j]==2 )
112                     D.add_edge( idx[i][j], D.dst, oo );
113             }
114         printf( "Case %d:\n%d\n", cas, D.maxflow() );
115     }
116 }
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posted @ 2015-03-07 18:01  idy002  阅读(110)  评论(0编辑  收藏  举报