HDU 3605 Escape（网络流 + 状压简化 ）

题意：

有 n 个人选择 m 个星球居住，选择情况和星球居住上限有限制，问是否能全部满足要求。

解题思路：

开始想到了最裸的建图，果然 TLE 了。

由于 n 太大（1e5） m 很小（10），所以我们可以进行状态压缩，将 n 个人等价划分成 1 << m 个集合，记录下各个集合有多少人。

建边：S -> mask，cap = cnt[mask] ； mask -> planet of mask，cap = inf；planet -> T，cap = limit of planet。

#include <set>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <cmath>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define L(i) i<<1
#define R(i) i<<1|1
#define INF  0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-9
#define maxn 100100
#define MOD 1000000007

const int MAXN = 100010;
const int MAXM = 400010;
struct Edge
{
    int to,next,cap,flow;
} edge[MAXM];
int tot,n,m;
int head[MAXN];
int gap[MAXN],dep[MAXN];
int pre[MAXN],cur[MAXN];

void init()
{
    tot = 0;
    memset(head,-1,sizeof(head));
}
void add_edge(int u,int v,int w,int rw = 0)
{
    edge[tot].to = v;
    edge[tot].cap = w;
    edge[tot].flow = 0;
    edge[tot].next = head[u];
    head[u] = tot++;
    edge[tot].to = u;
    edge[tot].cap = rw;
    edge[tot].flow = 0;
    edge[tot].next = head[v];
    head[v] = tot++;
}
int sap(int start,int en,int N)
{
    memset(gap,0,sizeof(gap));
    memset(dep,0,sizeof(dep));
    memcpy(cur,head,sizeof(head));
    int u = start;
    pre[u] = -1;
    gap[0] = N;
    int ans = 0;
    while(dep[start] < N)
    {
        if(u == en)
        {
            int Min = INF;
            for(int i = pre[u]; i != -1; i = pre[edge[i^1].to])
                if(Min > edge[i].cap - edge[i].flow)
                    Min = edge[i].cap - edge[i].flow;
            for(int i = pre[u]; i != -1; i = pre[edge[i^1].to])
            {
                edge[i].flow += Min;
                edge[i^1].flow -= Min;
            }
            u = start;
            ans += Min;
            continue;
        }
        int flag = 0;
        int v;
        for(int i = cur[u]; i != -1; i = edge[i].next)
        {
            v = edge[i].to;
            if(edge[i].cap - edge[i].flow && dep[v]+1 == dep[u])
            {
                flag = 1;
                cur[u] = pre[v] = i;
                break;
            }
        }
        if(flag)
        {
            u = v;
            continue;
        }
        int Min = N;
        for(int i = head[u]; i != -1; i = edge[i].next)
            if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min)
            {
                Min = dep[edge[i].to];
                cur[u] = i;
            }
        gap[dep[u]]--;
        if(!gap[dep[u]])
            return ans;
        dep[u] = Min + 1;
        gap[dep[u]]++;
        if(u != start)
            u = edge[pre[u]^1].to;
    }
    return ans;
}
int num[1050];

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int t,C = 1;
    //scanf("%d",&t);
    while(scanf("%d%d",&n,&m) != EOF)
    {
        init();
        memset(num,0,sizeof(num));
        for(int i = 1; i <= n; i++)
        {
            int sum = 0;
            for(int j = 0; j < m; j++)
            {
                int x;
                scanf("%d",&x);
                sum = sum * 2 + x;
            }
            num[sum]++;
        }
        for(int i = 0; i < (1<<m); i++)
        {
            add_edge(0,i+1,num[i]);
            int x = i,pos = 1;
            while(x)
            {
                if(x&1)
                    add_edge(i+1,pos+(1<<m),INF);
                x >>= 1;
                pos++;
            }
        }
        for(int j = 1; j <= m; j++)
        {
            int x;
            scanf("%d",&x);
            add_edge(j+(1<<m),(1<<m)+m+1,x);
        }
        printf("%s\n",sap(0,(1<<m)+m+1,(1<<m)+m+2) == n?"YES":"NO");
    }
    return 0;
}