bzoj2132【圈地计划】

题面

思路：

一开始以为和为了博多一样，两边连一样的，后来发现中间连负边的话根本不会割，即割断两块收益为负，所以WA的起飞……

正解是先黑白染色，每个点和它周围的点连边方式不同。对于黑点A，S-->A表示商业区的价值，A-->T表示工业区的价值，白点相反，对于相邻的情况，边权表示相邻的价值和，这样割断相当于选择同一用途，代价为正。

  1 #include <map>
  2 #include <set>
  3 #include <cmath>
  4 #include <queue>
  5 #include <stack>
  6 #include <cstdio>
  7 #include <string>
  8 #include <vector>
  9 #include <cstring>
 10 #include <complex>
 11 #include <cstdlib>
 12 #include <iostream>
 13 #include <algorithm>
 14 #define rg register
 15 #define ll long long
 16 using namespace std;
 17
 18 inline int gi()
 19 {
 20     rg int r = 0; rg bool b = 1; rg char c = getchar();
 21     while ((c < ‘0‘ || c > ‘9‘) && c != ‘-‘) c = getchar();
 22     if (c == ‘-‘) b = 0;
 23     while (c >= ‘0‘ && c <= ‘9‘) { r = (r << 1) + (r << 3) + c - ‘0‘, c = getchar(); }
 24     if (b) return r; return -r;
 25 }
 26
 27 const int inf = 2e8+5, N = 105, M = 5e5+1;
 28 int n,m,num,S,T;
 29 int dx[]={0,0,-1,1},dy[]={-1,1,0,0},dep[N*N],f[N*N],c[N][N];
 30 bool col[N][N];
 31 queue <int> q;
 32 struct Edge
 33 {
 34     int to,nx,cap;
 35 }eg[M];
 36
 37 inline void add(int x,int y,int z)
 38 {
 39     eg[++num]=(Edge){y,f[x],z}, f[x]=num;
 40 }
 41
 42 inline void link(int x,int y,int z)
 43 {
 44     add(x,y,z), add(y,x,0);
 45 }
 46
 47 inline int dfs(int o,int flow)
 48 {
 49     if (o == T || !flow) return flow;
 50     rg int to,cap,fl,tmp,i;
 51     fl=0;
 52     for (i=f[o]; i; i=eg[i].nx)
 53         {
 54             to=eg[i].to, cap=eg[i].cap;
 55             if (dep[o] != dep[to]-1 || cap <= 0)
 56                 continue;
 57             tmp=dfs(to,min(flow,cap));
 58             if (!tmp) continue;
 59             fl+=tmp, flow-=tmp;
 60             eg[i].cap-=tmp, eg[i^1].cap+=tmp;
 61             if (!flow) break;
 62         }
 63     if (!fl) dep[o]=0;
 64     return fl;
 65 }
 66
 67 inline int bfs()
 68 {
 69     memset(dep,0,sizeof(dep));
 70     rg int o,to,cap,i;
 71     q.push(S); dep[S]=1;
 72     while (!q.empty())
 73         {
 74             o=q.front(), q.pop();
 75             for (i=f[o]; i; i=eg[i].nx)
 76                 {
 77                     to=eg[i].to, cap=eg[i].cap;
 78                     if (dep[to] || cap <= 0)
 79                         continue;
 80                     dep[to]=dep[o]+1, q.push(to);
 81                 }
 82         }
 83     return dep[T];
 84 }
 85
 86 inline int dinic()
 87 {
 88     rg int flow;
 89     flow=0;
 90     while (bfs()) flow+=dfs(S,inf);
 91     return flow;
 92 }
 93
 94 int main()
 95 {
 96     rg int i,j,k,x,y,dis,ans;
 97     n=gi(), m=gi();
 98     ans=S=0, T=n*m+1, num=1;
 99     for (i=1; i<=n; ++i)
100         for (j=1; j<=m; ++j)
101             if ((i+j)&1)
102                 col[i][j]=1;
103     for (i=1; i<=n; ++i)
104         for (j=1; j<=m; ++j)
105             {
106                 dis=gi(), ans+=dis;
107                 if (col[i][j])
108                     link(S,(i-1)*m+j,dis);
109                 else
110                     link((i-1)*m+j,T,dis);
111             }
112
113     for (i=1; i<=n; ++i)
114         for (j=1; j<=m; ++j)
115             {
116                 dis=gi(), ans+=dis;
117                 if (col[i][j])
118                     link((i-1)*m+j,T,dis);
119                 else
120                     link(S,(i-1)*m+j,dis);
121             }
122     for (i=1; i<=n; ++i) for (j=1; j<=m; ++j) c[i][j]=gi();
123     for (i=1; i<=n;++i)
124         for (j=1; j<=m; ++j)
125             {
126                 if (!col[i][j]) continue;
127                 for (k=0; k<4; ++k)
128                     {
129                         x=i+dx[k], y=j+dy[k];
130                         if (x > n || x < 1 || y > m || y < 1)
131                             continue;
132                         add((i-1)*m+j,(x-1)*m+y,c[i][j]+c[x][y]);
133                         add((x-1)*m+y,(i-1)*m+j,c[i][j]+c[x][y]);
134                         ans+=c[i][j]+c[x][y];
135                     }
136             }
137     printf("%d\n",ans-dinic());
138     return 0;
139 }