T_T我太弱了这个题都不会做。
反向考虑:
可以把最小问题转化为最大问题:如果把格子填满,我们最多能拿走几个?
这样做法就很显然了。。行列建图就好了。。qwq
有时候脑子需要拐个弯T_T
#include<iostream>
#include<cstdio>
#include<cstring>
#define inf 1000000007
using namespace std;
int n,m,k,sum,cnt=1,S,T,a[105][105],l[105],c[105],sumx[105],sumy[105];
int head[205],cur[205],dis[205],q[205];
int next[100005],list[100005],key[100005];
inline int read()
{
int a=0,f=1; char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1; c=getchar();}
while (c>='0'&&c<='9') {a=a*10+c-'0'; c=getchar();}
return a*f;
}
inline void insert(int x,int y,int z)
{
next[++cnt]=head[x];
head[x]=cnt;
list[cnt]=y;
key[cnt]=z;
}
inline bool BFS()
{
memset(dis,-1,sizeof(dis));
dis[S]=1; q[1]=S;
int t=0,w=1,x;
while (t<w)
{
x=q[++t];
for (int i=head[x];i;i=next[i])
if (key[i]&&dis[list[i]]==-1)
dis[list[i]]=dis[x]+1,q[++w]=list[i];
}
return dis[T]!=-1;
}
int find(int x,int flow)
{
if (x==T) return flow;
int w,used=0;
for (int i=cur[x];i;i=next[i])
if (key[i]&&dis[list[i]]==dis[x]+1)
{
w=find(list[i],min(key[i],flow-used));
key[i]-=w; key[i^1]+=w; used+=w;
if (key[i]) cur[x]=i;
if (used==flow) return flow;
}
if (!used) dis[x]=-1;
return used;
}
inline int dinic()
{
int tmp=0;
while (BFS())
{
for (int i=S;i<=T;i++) cur[i]=head[i];
tmp+=find(S,inf);
}
return tmp;
}
int main()
{
n=read(); m=read(); k=read();
for (int i=1;i<=n;i++) l[i]=m-read();
for (int i=1;i<=m;i++) c[i]=n-read();
for (int i=1;i<=k;i++)
{
int x=read(),y=read();
sumx[x]++; sumy[y]++;
a[x][y]=1;
if (sumy[x]>l[x]||sumy[y]>c[y]) {puts("JIONG!"); return 0;}
}
sum=n*m-k; S=0; T=n+m+1;
for (int i=1;i<=n;i++) insert(S,i,l[i]-sumx[i]),insert(i,S,0);
for (int i=1;i<=m;i++) insert(i+n,T,c[i]-sumy[i]),insert(T,i+n,0);
for (int i=1;i<=n;i++)
for (int j=1;j<=m;j++)
if (!a[i][j]) insert(i,j+n,1),insert(j+n,i,0);
cout << sum-dinic() << endl;
return 0;
}