有一个N×M的单位方格中，其中有些方格是水塘，其他方格是陆地。如果要用1×2的矩阵区覆盖（覆盖过程不容许有任何部分重叠）这个陆地，那么最多可以覆盖多少陆地面积。
分析：一道很水的二分图匹配，染色染好后直接dinic求最大匹配数
代码：
#include
#include
#include
#include
#include
using namespace std;
int map[101][101];
int n,g,m;
const
int
INF=1<<30;
inthead[20002],next[50002],end,start=0,from[50002],to[50002],cap[50002],flow[52000];
bool is_start(int x,int y)
{
return 0==(x-y)%2;//0 1
}
int sum,side=1;
void read()
{
string s1;
scanf("%d%d",&n,&m);
scanf("%d",&g);
inti,j,k,l;
for(i=1;i<=g;i++)
{
scanf("%d%d",&j,&k);
map[j][k]=-1;
}
for(i=1;i<=n;i++)
for(j=1;j<=m;j++)
if(map[i][j]!=-1)
map[i][j]=++sum;
end=++sum;
for(i=1;i<=n;i++)
for(j=1;j<=m;j++)
{
if(map[i][j]!=-1)
{
if(is_start(i,j))
{
next[++side]=head[0];
head[0]=side;
to[side]=map[i][j];
from[side]=0;
cap[side]=1;
next[++side]=head[map[i][j]];
head[map[i][j]]=side;
to[side]=0;
from[side]=map[i][j];
cap[side]=0;
if(i+1<=n&&map[i+1][j]!=-1)
{
next[++side]=head[map[i][j]];
head[map[i][j]]=side;
to[side]=map[i+1][j];
from[side]=map[i][j];
cap[side]=1;
next[++side]=head[map[i+1][j]];
head[map[i+1][j]]=side;
to[side]=map[i][j];
from[side]=map[i+1][j];
cap[side]=0;
}
if(i-1>=1&&map[i-1][j]!=-1)
{
next[++side]=head[map[i][j]];
head[map[i][j]]=side;
to[side]=map[i-1][j];
from[side]=map[i][j];
cap[side]=1;
next[++side]=head[map[i-1][j]];
head[map[i-1][j]]=side;
to[side]=map[i][j];
from[side]=map[i-1][j];
cap[side]=0;
}
if(j+1<=m&&map[i][j+1]!=-1)
{
next[++side]=head[map[i][j]];
head[map[i][j]]=side;
to[side]=map[i][j+1];
from[side]=map[i][j];
cap[side]=1;
next[++side]=head[map[i][j+1]];
head[map[i][j+1]]=side;
to[side]=map[i][j];
from[side]=map[i][j+1];
cap[side]=0;
}
if(j-1>=1&&map[i][j-1]!=-1)
{
next[++side]=head[map[i][j]];
head[map[i][j]]=side;
to[side]=map[i][j-1];
from[side]=map[i][j];
cap[side]=1;
next[++side]=head[map[i][j-1]];
head[map[i][j-1]]=side;
to[side]=map[i][j];
from[side]=map[i][j-1];
cap[side]=0;
}
}
else
{
next[++side]=head[map[i][j]];
head[map[i][j]]=side;
to[side]=end;
from[side]=map[i][j];
cap[side]=1;
next[++side]=head[end];
head[end]=side;
to[side]=map[i][j];
from[side]=end;
cap[side]=0;
}
}
}
}
int h[15001];
bool visited[15001];
int line [15001];
bool BFS()
{
int left,right,l,r,i,j,k;
visited[start]=true;
line[1]=start;
h[start]=1;
left=1;
right=1;
while(left<=right)
{
l=left%(sum+1);
r=right%(sum+1);
j=head[line[l]];
while(j!=0)
{
if(!visited[to[j]]&&cap[j]>flow[j])
{
h[to[j]]=h[line[l]]+1;
line[(++right)%(sum+1)]=to[j];
visited[to[j]]=true;
}
j=next[j];
}
left++;
}
return visited[end];
}
int DFS(int x,int now)
{
int a=now;
int i,j,k,l;
j=head[x];
if(x==end||now==0)
returnnow;
while(j!=0)
{
if(cap[j]>flow[j]&&h[to[j]]==h[from[j]]+1)
{ l=DFS(to[j],min(a,cap[j]-flow[j])) ;
a-=l;
flow[j]+=l;
flow[j^1]-=l;
if(a==0)
return now;
}
j=next[j];
}
return now-a;
}
void find()
{
int ans=0;
while(BFS())
{
ans+=DFS(start,INF);
memset(visited,false,sizeof(visited));
memset(h,0,sizeof(h));
memset(line,0,sizeof(line));
}
printf("%d\n",ans);
}
int main()
{
int i,k;
read();
find();
return 0;
}
