HDU 3605 Escape（状压+最大流）

Escape

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 9430    Accepted Submission(s): 2234

Problem 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.

 

Sample Input

1 1

1

1

 

2 2

1 0

1 0

1 1

 

Sample Output

YES

NO

 

题目链接：HDU 3605

比较入门的一道最大流，显然可以想到是超级源点S->人连一条为流量1的边；人->可达的星球连一条流量为1的边；星球->超级汇点T连一条容量为给定的星球容量的边。但是这样就超时了。

膜了一下正确的做法，因为星球数不超过10，显然一个人的选择状态最多$2^{10}=1024$种，然后就把状态代替了人，因此上面的建图中把人改为二进制的状压转化为10进制的数值i即可，当然与人有关的流量也不再是1，而是同一状态下的人数

代码：

#include <stdio.h>
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int S=1050;
struct edge
{
    int to,nxt;
    int cap;
};
edge E[S*23];
int head[S],tot;
bitset<S> vis;
int st[S];

void init()
{
    CLR(head,-1);
    tot=0;
    CLR(st,0);
}
inline void add(int s,int t,int c)
{
    E[tot].to=t;
    E[tot].cap=c;
    E[tot].nxt=head[s];
    head[s]=tot++;

    E[tot].cap=0;
    E[tot].to=s;
    E[tot].nxt=head[t];
    head[t]=tot++;
}
int dfs(int s,int t,int f)
{
    if(s==t)
        return f;
    vis[s]=1;
    for (int i=head[s]; ~i; i=E[i].nxt)
    {
        int v=E[i].to;
        if(!vis[v]&&E[i].cap>0)
        {
            int d=dfs(v,t,min(f,E[i].cap));
            if(d>0)
            {
                E[i].cap-=d;
                E[i^1].cap+=d;
                return d;
            }
        }
    }
    return 0;
}
int max_flow(int s,int t)
{
    int ret=0;
    int f;
    while (true)
    {
        vis.reset();
        f=dfs(s,t,INF);
        if(!f)
            break;
        ret+=f;
    }
    return ret;
}
int main(void)
{
    int n,m,i,j,c,s;
    while (~scanf("%d%d",&n,&m))
    {
        init();
        int S=0;
        int T=(1<<m)+m;
        for (i=1; i<=n; ++i)
        {
            int S=0;
            for (j=0; j<m; ++j)
            {
                scanf("%d",&s);
                if(s)
                    S|=(1<<j);
            }
            ++st[S];
        }
        int R=1<<m;
        for (i=1; i<R; ++i)
        {
            if(st[i])
            {
                add(S,i,st[i]);
                for (j=0; j<m; ++j)
                {
                    if(i&(1<<j))
                        add(i,R+j,st[i]);
                }
            }
        }
        for (i=0; i<m; ++i)
        {
            scanf("%d",&c);
            add(R+i,T,c);
        }
        puts(max_flow(S,T)==n?"YES":"NO");
    }
    return 0;
}