hdu 3605 Escape (最大流+状态压缩)

Escape

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.

解题思路：因为题目中给的节点个数过多，如果之间建图连边会TLE，但是我们可以发现星球的个数非常的少，所以其实每个人对应各个星球的状态是可以合并的，也就是最多1024种情况，所以我们将这些节点状态压缩，然后建立一个超级源点，这个点连向那1024个新节点，流量上限是每个节点所对应的人数，这1024个新节点再按照2进制展开对10个星球连边，边权也是每个节点对应的人数，最后10个星球向汇点连边，边权是每个星球的容量。

AC代码：

/** @Author: wchhlbt* @Last Modified time: 2017-09-30*///#include <bits/stdc++.h>#include <vector>#include <list>#include <map>#include <set>#include <queue>#include <stack>#include <bitset>#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <sstream>#include <iostream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <ctime>#include <cstring>#include <limits>#include <climits>#include <cstdio>#define inf 0x3f3f3f3f#define pb push_back#define AA first#define BB second#define ONES(x) __builtin_popcount(x)#define _  << "  " <<using namespace std;typedef pair<int, int> P;typedef long long ll ;int dx[4] = {0,0,1,-1};int dy[4] = {1,-1,0,0};const double eps =1e-8;const int mod = 1000000007;const double PI = acos(-1.0);//inline int read(){ int num;    scanf("%d",&num);   return num;}const int maxn = 2000;inline int read(){    int data=0,w=1; char ch=0;    while(ch!='-' && (ch<'0' || ch>'9')) ch=getchar();    if(ch=='-') w=-1,ch=getchar();    while(ch>='0' && ch<='9') data=data*10+ch-'0',ch=getchar();    return data*w;}/*最大流Dinic算法使用前调用init函数*/struct Edge{    int from,to,cap,flow;    Edge(){}    Edge(int f,int t,int c,int fl):from(f),to(t),cap(c),flow(fl){}};struct Dinic{    int n,m,s,t;    vector<Edge> edges;    vector<int> G[maxn];    bool vis[maxn];    int cur[maxn];    int d[maxn];    void init(int n,int s,int t)//n是总节点数    {        this->n=n, this->s=s, this->t=t;        edges.clear();        for(int i=0;i<n;++i) G[i].clear();    }    void AddEdge(int from,int to,int cap)    {        edges.push_back( Edge(from,to,cap,0) );        edges.push_back( Edge(to,from,0,0) );        m=edges.size();        G[from].push_back(m-2);        G[to].push_back(m-1);    }    bool BFS()    {        queue<int> Q;        memset(vis,0,sizeof(vis));        vis[s]=true;        d[s]=0;        Q.push(s);        while(!Q.empty())        {            int x=Q.front(); Q.pop();            for(int i=0;i<G[x].size();++i)            {                Edge& e=edges[G[x][i]];                if(!vis[e.to] && e.cap>e.flow)                {                    vis[e.to]=true;                    d[e.to]=d[x]+1;                    Q.push(e.to);                }            }        }        return vis[t];    }    int DFS(int x,int a)    {        if(x==t || a==0) return a;        int flow=0, f;        for(int &i=cur[x];i<G[x].size();++i)        {            Edge &e=edges[G[x][i]];            if(d[e.to]==d[x]+1 && (f=DFS(e.to,min(a,e.cap-e.flow) ) )>0)            {                e.flow +=f;                edges[G[x][i]^1].flow -=f;                flow +=f;                a -=f;                if(a==0) break;            }        }        return flow;    }    int max_flow()    {        int ans=0;        while(BFS())        {            memset(cur,0,sizeof(cur));            ans +=DFS(s,inf);        }        return ans;    }}dinic;int a[1200];int b[20];int main(){int n,m;while(~scanf("%d%d",&n,&m)){dinic.init(1026+m,0,1025+m);memset(a,0,sizeof a);for(int i = 1; i<=n; i++){int num = 0;for(int j = 1; j<=m; j++){//2进制压缩int t = read();num *= 2;if(t)num += 1;}a[num+1]++;}for(int i = 1; i<=1024; i++){if(a[i]!=0){dinic.AddEdge(0,i,a[i]);int t = i;t--;for(int j = m; j>=1; j--){//按照2进制位展开建图if(t%2)dinic.AddEdge(i,1024+j,a[i]);t /= 2;}}}for(int i = 1; i<=m; i++){b[i] = read();dinic.AddEdge(1024+i,1025+m,b[i]);}if(dinic.max_flow()==n)puts("YES");elseputs("NO");}    return 0;}