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;
}