HDU- 3605 - Escape 最大流 + 二进制压位

HDU - 3605 ： acm.hdu.edu.cn/showproblem.php?pid=3605

题目：

　　　　有1e5的的人，小于10个的星球，每个星球都有容量，每个人也有适合的星球和不适合的星球。问所有人是否能住到星球上去。

思路：

　　　　这道题目如果直接用常用的建图方式是不行的，因为人太多了，相应的边就很多，会TLE。那要怎么做呢，因为只有十个星球，所以有很多人的特征是相同的，所以把这1e5个点可以缩成最多1024个点。这样建图跑dinic就行了。

//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000")  //c++
// #pragma GCC diagnostic error "-std=c++11"
// #pragma comment(linker, "/stack:200000000")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
// #pragma GCC optimize("-fdelete-null-pointer-checks,inline-functions-called-once,-funsafe-loop-optimizations,-fexpensive-optimizations,-foptimize-sibling-calls,-ftree-switch-conversion,-finline-small-functions,inline-small-functions,-frerun-cse-after-loop,-fhoist-adjacent-loads,-findirect-inlining,-freorder-functions,no-stack-protector,-fpartial-inlining,-fsched-interblock,-fcse-follow-jumps,-fcse-skip-blocks,-falign-functions,-fstrict-overflow,-fstrict-aliasing,-fschedule-insns2,-ftree-tail-merge,inline-functions,-fschedule-insns,-freorder-blocks,-fwhole-program,-funroll-loops,-fthread-jumps,-fcrossjumping,-fcaller-saves,-fdevirtualize,-falign-labels,-falign-loops,-falign-jumps,unroll-loops,-fsched-spec,-ffast-math,Ofast,inline,-fgcse,-fgcse-lm,-fipa-sra,-ftree-pre,-ftree-vrp,-fpeephole2",3)

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>

using namespace std;
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue



typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n'

#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
#define max3(a,b,c) max(max(a,b), c);
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1e9+7;
const double esp = 1e-8;
const double PI=acos(-1.0);
const double PHI=0.61803399;    //黄金分割点
const double tPHI=0.38196601;


template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x=f?-x:x;
}


/*-----------------------showtime----------------------*/

             struct edge{
                int u,v,cap,flag;
                edge(){}
                edge(int u,int v,int cap,int flag):u(u),v(v),cap(cap),flag(flag){}
            }es[100009];

            int tot,s,t;
            vector<int>tab[2009];
            int dis[2009],cur[2009];
            void addedge(int u,int v,int cap){
             //   debug(u);
                tab[u].pb(tot);
                es[tot++] = edge(u,v,cap,1);
                tab[v].pb(tot);
                es[tot++] = edge(v,u,0,0);
            }

            bool bfs(){
                queue<int>q; q.push(s);
                memset(dis,inf,sizeof(dis));
                dis[s] = 0;
                while(!q.empty()){
                    int h = q.front(); q.pop();
                    for(int i=0; i<tab[h].size(); i++){
                        edge & e = es[tab[h][i]];
                        if(e.cap > 0 && dis[e.v] >= inf){
                            dis[e.v] = dis[h] + 1;
                            q.push(e.v);
                        }
                    }
                }
                return dis[t] < inf;
            }

            int dfs(int x,int maxflow){
                if(x == t || maxflow == 0) return maxflow;
                for(int i=cur[x] ; i<tab[x].size(); i++){
                    cur[x] = i;
                    edge & e = es[tab[x][i]];
                    if(dis[e.v] == dis[x] + 1 && e.cap > 0){
                        int flow = dfs(e.v, min(maxflow, e.cap));
                        if(flow){
                            e.cap -= flow;

                            es[tab[x][i] ^ 1].cap += flow;
                            return flow;
                        }
                    }
                }
                return 0;
            }

            int dinic(){
                int ans = 0;
                while(bfs()){

                    int flow;
                    memset(cur,0,sizeof(cur));
                    do{
                        flow = dfs(s,inf);
                        if(flow) ans += flow;
                    }while(flow);

                }
                return ans;
            }
            int cnt[2000];
int main(){
            int n,m;
            while(~scanf("%d%d", &n, &m) && n+m){

                tot = 0;
                s = 0, t = 1024+m+m+1;
                for(int i=s; i<=t; i++)tab[i].clear();
                memset(cnt,0,sizeof(cnt));
                for(int i=1; i<=n; i++){
                    int tmp = 0;
                    for(int j=1; j<=m; j++){
                        int x;      scanf("%d", &x);
                        tmp= tmp*2 + x;
                    }
                    cnt[tmp]++;
                }
                for(int i=1; i<=1024; i++)
                {
                    if(cnt[i] == 0)continue;
                    addedge(s,i,cnt[i]);
                    for(int j=0; j<m; j++){
                        if((i>>j) & 1 == 1) addedge(i,1025+j,cnt[i]);
                    }
                }
                for(int i=0; i<m; i++){
                    int x;scanf("%d", &x);
                    addedge(1025+i,1025+m+i,x);
                    addedge(1025+m+i,t,inf);
                }
                if(n == dinic()) puts("YES");
                else puts("NO");
            }
            return 0;
}