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。

#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <stack>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <complex>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define rg register
#define ll long long
using namespace std;

inline int gi()
{
    rg int r = 0; rg bool b = 1; rg char c = getchar();
    while ((c < '0' || c > '9') && c != '-') c = getchar();
    if (c == '-') b = 0;
    while (c >= '0' && c <= '9') { r = (r << 1) + (r << 3) + c - '0', c = getchar(); }
    if (b) return r; return -r;
}

const int inf = 2e8+5, N = 105, M = 5e5+1;
int n,m,num,S,T;
int dx[]={0,0,-1,1},dy[]={-1,1,0,0},dep[N*N],f[N*N],c[N][N];
bool col[N][N];
queue <int> q;
struct Edge
{
    int to,nx,cap;
}eg[M];

inline void add(int x,int y,int z)
{
    eg[++num]=(Edge){y,f[x],z}, f[x]=num;
}

inline void link(int x,int y,int z)
{
    add(x,y,z), add(y,x,0);
}

inline int dfs(int o,int flow)
{
    if (o == T || !flow) return flow;
    rg int to,cap,fl,tmp,i;
    fl=0;
    for (i=f[o]; i; i=eg[i].nx)
        {
            to=eg[i].to, cap=eg[i].cap;
            if (dep[o] != dep[to]-1 || cap <= 0)
                continue;
            tmp=dfs(to,min(flow,cap));
            if (!tmp) continue;
            fl+=tmp, flow-=tmp;
            eg[i].cap-=tmp, eg[i^1].cap+=tmp;
            if (!flow) break;
        }
    if (!fl) dep[o]=0;
    return fl;
}

inline int bfs()
{
    memset(dep,0,sizeof(dep));
    rg int o,to,cap,i;
    q.push(S); dep[S]=1;
    while (!q.empty())
        {
            o=q.front(), q.pop();
            for (i=f[o]; i; i=eg[i].nx)
                {
                    to=eg[i].to, cap=eg[i].cap;
                    if (dep[to] || cap <= 0)
                        continue;
                    dep[to]=dep[o]+1, q.push(to);
                }
        }
    return dep[T];
}

inline int dinic()
{
    rg int flow;
    flow=0;
    while (bfs()) flow+=dfs(S,inf);
    return flow;
}

int main()
{
    rg int i,j,k,x,y,dis,ans;
    n=gi(), m=gi();
    ans=S=0, T=n*m+1, num=1;
    for (i=1; i<=n; ++i)
        for (j=1; j<=m; ++j)
            if ((i+j)&1)
                col[i][j]=1;
    for (i=1; i<=n; ++i)
        for (j=1; j<=m; ++j)
            {
                dis=gi(), ans+=dis;
                if (col[i][j])
                    link(S,(i-1)*m+j,dis);
                else
                    link((i-1)*m+j,T,dis);
            }

    for (i=1; i<=n; ++i)
        for (j=1; j<=m; ++j)
            {
                dis=gi(), ans+=dis;
                if (col[i][j])
                    link((i-1)*m+j,T,dis);
                else
                    link(S,(i-1)*m+j,dis);
            }
    for (i=1; i<=n; ++i) for (j=1; j<=m; ++j) c[i][j]=gi();
    for (i=1; i<=n;++i)
        for (j=1; j<=m; ++j)
            {
                if (!col[i][j]) continue;
                for (k=0; k<4; ++k)
                    {
                        x=i+dx[k], y=j+dy[k];
                        if (x > n || x < 1 || y > m || y < 1)
                            continue;
                        add((i-1)*m+j,(x-1)*m+y,c[i][j]+c[x][y]);
                        add((x-1)*m+y,(i-1)*m+j,c[i][j]+c[x][y]);
                        ans+=c[i][j]+c[x][y];
                    }
            }
    printf("%d
",ans-dinic());
    return 0;
}