最大流-HDU-3605-Escape

Escape

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 8376 Accepted Submission(s): 1893

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

题意:

一共有n个人，m个星球，每个人都有自己适应居住的星球，每个星球有自己容纳上限，现在问是否可以让所有人都住到这m个星球上去。

题解：

因为n非常大（1E5），而m又非常小（10），所以可以发现，人只会有2^m种情况，那么进行压缩后，先统计每种情况的人数，再从每种情况建网跑最大流就可以了。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <queue>
#include <vector>
#include <set>
#include <map>
#include <utility>
using namespace std;
const int MAXN = 2048;//点数的最大值
const int MAXM = 400010;//边数的最大值
const int INF = 0x3f3f3f3f;
struct Edge
{
int to,next,cap,flow;
} edge[MAXM]; //注意是MAXM
int tol;
int head[MAXN];
int gap[MAXN],dep[MAXN],cur[MAXN];
void init()
{
tol = 0;
memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int w,int rw = 0)
{
edge[tol].to = v;
edge[tol].cap = w;
edge[tol].flow = 0;
edge[tol].next = head[u];
head[u] = tol++;
edge[tol].to = u;
edge[tol].cap = rw;
edge[tol].flow = 0;
edge[tol].next = head[v];
head[v] = tol++;
}
int Q[MAXN];
void BFS(int start,int end)
{
memset(dep,-1,sizeof(dep));
memset(gap,0,sizeof(gap));
gap[0] = 1;
int front = 0, rear = 0;
dep[end] = 0;
Q[rear++] = end;
while(front != rear)
{
int u = Q[front++];
for(int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if(dep[v] != -1)continue;
Q[rear++] = v;
dep[v] = dep[u] + 1;
gap[dep[v]]++;
}
}
}
int S[MAXN];
int sap(int start,int end,int N)
{
BFS(start,end);
memcpy(cur,head,sizeof(head));
int top = 0;
int u = start;
int ans = 0;
while(dep[start] < N)
{
if(u == end)
{
int Min = INF;
int inser;
for(int i = 0; i < top; i++)
if(Min > edge[S[i]].cap - edge[S[i]].flow)
{
Min = edge[S[i]].cap - edge[S[i]].flow;
inser = i;
}
for(int i = 0; i < top; i++)
{
edge[S[i]].flow += Min;
edge[S[i]^1].flow -= Min;
}
ans += Min;
top = inser;
u = edge[S[top]^1].to;
continue;
}
bool flag = false;
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 = true;
cur[u] = i;
break;
}
}
if(flag)
{
S[top++] = cur[u];
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[S[--top]^1].to;
}
return ans;
}
long long int n,num[1<<11];
int m,total;
int main()
{
int tmp,tmp1;
while(scanf("%lld%d",&n,&m)!=EOF)
{
init();
total=1<<m;
memset(num,0,sizeof(num));
for(long long int i=1;i<=n;i++)
{
tmp1=0;
for(int j=0;j<m;j++)
{
scanf("%d",&tmp);
if(tmp)
tmp1+=(1<<j);
}
num[tmp1]++;
}
for(long long int i=0;i<total;i++)
{
addedge(total+m,i,num[i]);
for(int j=0;j<m;j++)
{
if(i&(1<<j))
addedge(i,total+j,num[i]);
}
}
for(int i=0;i<m;i++)
{
scanf("%d",&tmp);
addedge(total+i,total+m+1,tmp);
}
if(sap(total+m,total+m+1,total+m+2)==n)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}