# HDU 3605 Escape （网络流，最大流，位运算压缩）

### Description

2012 If this is the end of the world how to do? I do not know how. But now scientists have found that some stars, who can live, but some people do not fit to live some of the planet. Now scientists want your help, is to determine what all of people can live in these planets.

### Input

More set of test data, the beginning of each data is n (1 <= n <= 100000), m (1 <= m <= 10) n indicate there n people on the earth, m representatives m planet, planet and people labels are from 0. Here are n lines, each line represents a suitable living conditions of people, each row has m digits, the ith digits is 1, said that a person is fit to live in the ith-planet, or is 0 for this person is not suitable for living in the ith planet.
The last line has m digits, the ith digit ai indicates the ith planet can contain ai people most..
0 <= ai <= 100000

### Output

Determine whether all people can live up to these stars
If you can output YES, otherwise output NO.

1 1
1
1

2 2
1 0
1 0
1 1

YES
NO

## 代码

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<map>
using namespace std;

const int maxN=100101;
const int maxM=maxN*70;
const int inf=2147483647;

class Edge
{
public:
int u,v,flow;
};

int n,m;
int cnt;
int tot;
int Next[maxM];
Edge E[maxM];
int depth[maxN];
int cur[maxN];
int Q[maxN];
map<int,int> Cnt;

bool bfs();
int dfs(int u,int flow);

int main()
{
while (cin>>n>>m)
{
cnt=-1;
tot=0;
Cnt.clear();
for (int i=1;i<=n;i++)
{
int ret=0;
for (int j=1;j<=m;j++)
{
int is;
scanf("%d",&is);
ret=ret+(is<<j);//位运算统计
}
Cnt[ret]++;//Cnt就是统计某一种类型的人有多少个
}
map<int,int>::iterator loop;
for (loop=Cnt.begin();loop!=Cnt.end();loop++)//遍历所有的类型
{
tot++;
for (int i=1;i<=m;i++)//连人与星球
if ((loop->first)&(1<<i))//若这类人都能适应第j个星球，则连边，注意容量为这类人的个数
}
for (int i=1;i<=m;i++)
{
int flow;
scanf("%d",&flow);//读入每一个星球的人数上限，并连汇点
}
int Ans=0;//求解最大流
while (bfs())
{
for (int i=0;i<=tot+m+1;i++)
while (int di=dfs(0,inf))
Ans+=di;
}
if (Ans==n)//若满流，则有解，否则无解
printf("YES\n");
else
printf("NO\n");
}
return 0;
}

{
cnt++;
E[cnt].u=u;
E[cnt].v=v;
E[cnt].flow=flow;

cnt++;
E[cnt].u=v;
E[cnt].v=u;
E[cnt].flow=0;
}

bool bfs()
{
memset(depth,-1,sizeof(depth));
int h=1,t=0;
Q[1]=0;
depth[0]=1;
do
{
t++;
int u=Q[t];
{
int v=E[i].v;
if ((depth[v]==-1)&&(E[i].flow>0))
{
depth[v]=depth[u]+1;
h++;
Q[h]=v;
}
}
}
while (h!=t);
if (depth[tot+m+1]==-1)
return 0;
return 1;
}

int dfs(int u,int flow)
{
if (u==tot+m+1)
return flow;
for (int &i=cur[u];i!=-1;i=Next[i])
{
int v=E[i].v;
if ((depth[v]==depth[u]+1)&&(E[i].flow>0))
{
int di=dfs(v,min(flow,E[i].flow));
if (di>0)
{
E[i].flow-=di;
E[i^1].flow+=di;
return di;
}
}
}
return 0;
}


posted @ 2017-08-20 16:18  SYCstudio  阅读(...)  评论(...编辑  收藏