Escape
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 12757 Accepted Submission(s): 3169
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
Source
2010 ACM-ICPC Multi-University Training Contest（17）——Host by ZSTU
因为m只有10，所以二进制将人分类最多2^10 = 1024种，然后建图跑最大流板子即可！
//Max_flow
//@2018/05/04 Friday
//SAP O(n^2 * m) O(m*3*2)
//by Tawn
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cstring>
#include <map>
#include <cmath>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 1000 + 100;
const int maxm = 1000000 + 10;
typedef pair<int,int> P;
int head[maxn];//链式前向星
int tot = 0,num;
int n,m;
struct edge
{
int to;
int c;
int next;
edge(int x = 0, int y = 0, int z = 0) : to(x), c(y), next(z) {}
}es[maxm*2];//记录边 注意是2倍
void add_edge(int u, int v, int c)
{
es[tot] = edge(v,c,head[u]);
head[u] = tot++;
}
int SAP(int s, int t)
{
int numh[maxn],h[maxn],ce[maxn],pre[maxn];
//numh 记录gap优化的统计高度数量数组，h 距离标号数组，ce 当前弧，pre前驱数组
int f, ans = 0, u, temp, neck, i; //初始化最大流为0
memset(h,0,sizeof(h));
memset(numh,0,sizeof(numh));
memset(pre,-1,sizeof(pre));
for(i = 0; i <= num; i++) ce[i] = head[i];
numh[0] = num;
u = s;
while(h[s] < num)
{
//寻找增广路
if(u == t)
{
f = INF;
for(i = s; i != t; i = es[ce[i]].to)
{
if(f > es[ce[i]].c)
{
neck = i;
f = es[ce[i]].c;
}
}
for(i = s; i != t; i = es[ce[i]].to)
{
temp = ce[i];
es[temp].c -= f;
es[temp^1].c += f;
}
ans += f;
u = neck;
}
//寻找可行弧
for(i = ce[u]; i != -1; i = es[i].next)
if(es[i].c && h[u] == h[es[i].to] + 1) break;
//寻找增广路
if(i != -1)
{
ce[u] = i;
pre[es[i].to] = u;
u = es[i].to;
}
else
{
if(!--numh[h[u]]) break; //gap optimization
ce[u] = head[u];
for(temp = num, i = head[u]; i != -1; i = es[i].next)
if(es[i].c) temp = min(temp, h[es[i].to]);
h[u] = temp + 1;
++numh[h[u]];
if(u != s) u = pre[u];//重标号并且从当前点前驱重新增广
}
}
return ans;
}
int a[1050];
int main()
{
while(~scanf("%d%d",&n,&m))
{
memset(head,-1,sizeof(head));
memset(a,0,sizeof(a));
int s = pow(2,m);
int t = s+m+1;
for(int i = 0; i < n; i++)
{
int res = 0,temp;
for(int j = 0; j < m; j++)
{
scanf("%d",&temp);
res *= 2;
res += temp;
}
a[res] ++;
}
//for(int i = 0; i < s; i++) cout << a[i] << " ";
//cout << endl;
for(int i = 0; i < s; i++)
{
if(a[i] != 0)
{
add_edge(s,i,a[i]);
add_edge(i,s,0);
int p = 1;
for(int x = 1; x < s; x <<= 1,p++) if(x & i) {add_edge(i,s+p,a[i]);add_edge(s+p,i,0); }
}
}
for(int i = m; i >= 1; i--)
{
int p;
scanf("%d",&p);
add_edge(s+i,t,p);
add_edge(t,s+i,0);
}
num = t + 1;
int ans = SAP(s,t);
if(ans == n) printf("YES\n");
else printf("NO\n");
// cout << s << " " << t << " " << ans << endl;
}
return 0;
}