HDU 3065 Escape(简单网络流 二分图多重匹配）

#include <iostream>
#include <cstdio>
#include <climits>
#include <cstring>
#include <algorithm>
using namespace std;
typedef  struct {int v,next,val;} edge;
const int MAXN=2100;
const int MAXM=23000;
edge e[MAXM];
int p[MAXN],eid;
int map[MAXN];
inline void init(){memset(p,-1,sizeof(p));eid=0;}
//有向
inline void insert1(int from,int to,int val)
{
e[eid].v=to;e[eid].val=val;
e[eid].next=p[from];
p[from]=eid++;
swap(from,to);
e[eid].v=to;e[eid].val=0;
e[eid].next=p[from];
p[from]=eid++;
}
//无向
inline void insert2(int from,int to,int val)
{
e[eid].v=to;e[eid].val=val;
e[eid].next=p[from];
p[from]=eid++;
swap(from,to);
e[eid].v=to;e[eid].val=val;
e[eid].next=p[from];
p[from]=eid++;
}
int n,m;//n为点数 m为边数
int h[MAXN];
int gap[MAXN];
int source,sink;
inline int dfs(int pos,int cost)
{
if (pos==sink) return cost;

int j,minh=n-1,lv=cost,d;
for (j=p[pos];j!=-1;j=e[j].next)
{
int v=e[j].v,val=e[j].val;
if(val>0)
{
if (h[v]+1==h[pos])
{
if (lv<e[j].val) d=lv;
else d=e[j].val;

d=dfs(v,d);
e[j].val-=d;
e[j^1].val+=d;
lv-=d;
if (h[source]>=n) return cost-lv;
if (lv==0) break;
}
if (h[v]<minh)    minh=h[v];
}
}
if (lv==cost)
{
--gap[h[pos]];
if (gap[h[pos]]==0) h[source]=n;
h[pos]=minh+1;
++gap[h[pos]];
}
return cost-lv;
}

int isap(int st,int ed)
{

source=st;sink=ed;
int ret=0;
memset(gap,0,sizeof(gap));
memset(h,0,sizeof(h));
gap[st]=n;
while (h[st]<n)
{
ret+=dfs(st,INT_MAX);
}
return ret;
}
int main()
{
int p,k,t,i,j,q,s;
while(scanf("%d%d",&p,&k)!=EOF)
{
init();
m=0;
memset(map,0,sizeof(map));
for(i=1;i<=p;i++){
t=1;
s=0;
for(j=0;j<k;j++){
scanf("%d",&q);
s+=q*t;
t<<=1;
}
map[s]++;
}
t=(1<<(k));
for(i=2;i<=t+1;i++){
insert1(1,i,map[i-2]);
m++;
}
n=0;
n+=t+1;
for(i=0;i<t;i++){
for(j=0;j<k;j++){
insert1(i+2,t+2+j,map[i]*(i&(1<<(j))));
m++;
}
}
for(j=0;j<k;j++){
scanf("%d",&q);
insert1(t+2+j,t+k+2,q);
m++;
}
insert1(t+k+2,t+k+3,100000000);
m++;
n+=k+2;
t=isap(1,n);
if(t>=p)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}