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  • hdu 1569 最小割

    和HDU 1565是一道题,只是数据加强了,貌似轮廓线DP来不了了。

      1 #include <cstdio>
      2 #include <cstring>
      3 #include <queue>
      4 #include <vector>
      5 #define maxn 2510
      6 #define oo 0x3f3f3f3f
      7 using namespace std;
      8 
      9 struct Edge {
     10     int u, v, f;
     11     Edge( int u, int v, int f ):u(u),v(v),f(f){}
     12 };
     13 struct Dinic {
     14     int n, src, dst;
     15     vector<Edge> edge;
     16     vector<int> g[maxn];
     17     int dep[maxn], cur[maxn];
     18 
     19     void init( int n, int src, int dst ) {
     20         this->n = n;
     21         this->src = src;
     22         this->dst = dst;
     23         for( int u=1; u<=n; u++ ) 
     24             g[u].clear();
     25         edge.clear();
     26     }
     27     void add_edge( int u, int v, int f ) {
     28         g[u].push_back( edge.size() );
     29         edge.push_back( Edge(u,v,f) );
     30         g[v].push_back( edge.size() );
     31         edge.push_back( Edge(v,u,0) );
     32     }
     33     bool bfs() {
     34         queue<int> qu;
     35         memset( dep, 0, sizeof(dep) );
     36         qu.push( src );
     37         dep[src] = 1;
     38         while( !qu.empty() ) {
     39             int u=qu.front();
     40             qu.pop();
     41             for( int t=0; t<g[u].size(); t++ ) {
     42                 Edge &e = edge[g[u][t]];
     43                 if( e.f && !dep[e.v] ) {
     44                     dep[e.v] = dep[e.u]+1;
     45                     qu.push( e.v );
     46                 }
     47             }
     48         }
     49         return dep[dst];
     50     }
     51     int dfs( int u, int a ) {
     52         if( u==dst || a==0 ) return a;
     53         int remain=a, past=0, na;
     54         for( int &t=cur[u]; t<g[u].size(); t++ ) {
     55             Edge &e = edge[g[u][t]];
     56             Edge &ve = edge[g[u][t]^1];
     57             if( dep[e.v]==dep[e.u]+1 && e.f && (na=dfs(e.v,min(remain,e.f))) ) {
     58                 remain -= na;
     59                 past += na;
     60                 e.f -= na;
     61                 ve.f += na;
     62                 if( !remain ) break;
     63             }
     64         }
     65         return past;
     66     }
     67     int maxflow() {
     68         int flow = 0;
     69         while( bfs() ) {
     70             memset( cur, 0, sizeof(cur) );
     71             flow += dfs(src,oo);
     72         }
     73         return flow;
     74     }
     75 };
     76 
     77 int n, m;
     78 int idx[55][55], id_clock;
     79 int arr[55][55], sum;
     80 int di[4] = { 0, 0, +1, -1 };
     81 int dj[4] = { +1, -1, 0, 0 };
     82 Dinic D;
     83 
     84 int main() {
     85     while( scanf( "%d%d", &n, &m )==2 ) {
     86         id_clock = 0;
     87         sum = 0;
     88         for( int i=1; i<=n; i++ ) 
     89             for( int j=1; j<=m; j++ ) {
     90                 scanf( "%d", &arr[i][j] );
     91                 sum += arr[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                 if( (i+j)&1 ) {
     98                     D.add_edge( D.src, idx[i][j], arr[i][j] );
     99                     for( int d=0; d<4; d++ ) {
    100                         int ni = i+di[d];
    101                         int nj = j+dj[d];
    102                         if( 1<=ni&&ni<=n && 1<=nj&&nj<=m ) {
    103                             int u = idx[i][j];
    104                             int v = idx[ni][nj];
    105                             D.add_edge( u, v, oo );
    106                         }
    107                     }
    108                 } else {
    109                     D.add_edge( idx[i][j], D.dst, arr[i][j] );
    110                 }
    111             }
    112         printf( "%d
    ", sum-D.maxflow() );
    113     }
    114 }
    View Code
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  • 原文地址:https://www.cnblogs.com/idy002/p/4321320.html
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