BZOJ2668: [cqoi2012]交换棋子

  1 #include<cstdio>
  2   
  3 #include<cstdlib>
  4   
  5 #include<cmath>
  6   
  7 #include<cstring>
  8   
  9 #include<algorithm>
 10   
 11 #include<iostream>
 12   
 13 #include<vector>
 14   
 15 #include<map>
 16   
 17 #include<set>
 18   
 19 #include<queue>
 20   
 21 #include<string>
 22   
 23 #define inf 1000000000
 24   
 25 #define maxn 200000+5
 26   
 27 #define maxm 200000+5
 28   
 29 #define eps 1e-10
 30   
 31 #define ll long long
 32   
 33 #define pa pair<int,int>
 34   
 35 #define for0(i,n) for(int i=0;i<=(n);i++)
 36   
 37 #define for1(i,n) for(int i=1;i<=(n);i++)
 38   
 39 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
 40   
 41 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
 42   
 43 #define for4(i,x) for(int i=head[x],y;i;i=e[i].next)
 44   
 45 #define mod 1000000007
 46   
 47 using namespace std;
 48   
 49 inline int read()
 50   
 51 {
 52   
 53     int x=0,f=1;char ch=getchar();
 54   
 55     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
 56   
 57     while(ch>=‘0‘&&ch<=‘9‘){x=10*x+ch-‘0‘;ch=getchar();}
 58   
 59     return x*f;
 60   
 61 }
 62 int n,m,k,mincost,tot,s,t,head[maxn],d[maxn],from[2*maxm];
 63 bool v[maxn];
 64 queue<int>q;
 65 int a[100][100][3],num[100][100][3],cnt[3];
 66 struct edge{int from,go,next,v,c;}e[2*maxm];
 67 void add(int x,int y,int v,int c)
 68 {
 69     e[++tot]=(edge){x,y,head[x],v,c};head[x]=tot;
 70     e[++tot]=(edge){y,x,head[y],0,-c};head[y]=tot;
 71 }
 72 bool spfa()
 73 {
 74     for (int i=s;i<=t;i++){v[i]=0;d[i]=inf;}
 75     q.push(s);d[s]=0;v[s]=1;
 76     while(!q.empty())
 77     {
 78         int x=q.front();q.pop();v[x]=0;
 79         for (int i=head[x],y;i;i=e[i].next)
 80          if(e[i].v&&d[x]+e[i].c<d[y=e[i].go])
 81          {
 82             d[y]=d[x]+e[i].c;from[y]=i;
 83             if(!v[y]){v[y]=1;q.push(y);}
 84          }
 85     }
 86     return d[t]!=inf;
 87 }
 88 void mcf()
 89 {
 90     mincost=0;
 91     while(spfa())
 92     {
 93         int tmp=inf;
 94         for(int i=from[t];i;i=from[e[i].from]) tmp=min(tmp,e[i].v);
 95         mincost+=d[t]*tmp;
 96         for(int i=from[t];i;i=from[e[i].from]){e[i].v-=tmp;e[i^1].v+=tmp;}
 97     }
 98 }
 99 const int dx[8]={0,0,1,-1,1,1,-1,-1};
100 const int dy[8]={1,-1,0,0,-1,1,1,-1};
101   
102 int main()
103   
104 {
105     n=read();m=read();s=0;t=3*n*m+1;
106     for0(k,2)for1(i,n)for1(j,m)
107     {
108         char ch=getchar();
109         while(ch<‘0‘||ch>‘9‘)ch=getchar();
110         a[i][j][k]=ch-‘0‘;
111         //cout<<i<<‘ ‘<<j<<‘ ‘<<k<<‘ ‘<<a[i][j][k]<<endl;
112         num[i][j][k]=++tot;
113         cnt[k]+=a[i][j][k];
114     }
115     //cout<<tot<<‘ ‘<<s<<‘ ‘<<t<<endl;
116     tot=1;
117     for1(i,n)for1(j,m)
118     {
119         if(a[i][j][0]&&a[i][j][1])a[i][j][0]=a[i][j][1]=0;
120         if(a[i][j][0])
121         {
122             add(num[i][j][1],num[i][j][0],a[i][j][2]/2,0);
123             add(num[i][j][0],num[i][j][2],(a[i][j][2]+1)/2,0);
124             add(s,num[i][j][0],1,0);
125         }
126         else if(a[i][j][1])
127         {
128             add(num[i][j][1],num[i][j][0],(a[i][j][2]+1)/2,0);
129             add(num[i][j][0],num[i][j][2],a[i][j][2]/2,0);
130             add(num[i][j][0],t,1,0);
131         }
132         else
133         {
134             add(num[i][j][1],num[i][j][0],a[i][j][2]/2,0);
135             add(num[i][j][0],num[i][j][2],a[i][j][2]/2,0);
136         }
137         for0(k,7)
138         {
139             int x=i+dx[k],y=j+dy[k];
140             if(x<1||x>n||y<1||y>m)continue;
141             add(num[i][j][2],num[x][y][1],inf,1);
142         }
143     }
144     if(cnt[0]!=cnt[1]){printf("-1\n");return 0;}
145     mcf();
146     printf("%d\n",mincost);
147   
148     return 0;
149   
150 }