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HDU3605(KB11-M 状态压缩+最大流)

Escape

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 10920    Accepted Submission(s): 2630

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
 
由于n很大，直接建图会T，但是m很小，因此根据一个人在能否在m个星球上生存的状态，可以压缩为一个m位二进制数。最多有2^10种状态。
源点向每一种状态连边，容量为该状态的人数。
每个状态向该状态下能够生存的星球连边，容量为该状态的人数。
每个星球向汇点连边，容量为星球最多能承受的人数。
  1 //2017-08-24
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <iostream>
  5 #include <algorithm>
  6 #include <queue>
  7 
  8 using namespace std;
  9 
 10 const int N = 110000;
 11 const int M = 5100100;
 12 const int INF = 0x3f3f3f3f;
 13 int head[N], tot;
 14 struct Edge{
 15     int next, to, w;
 16 }edge[M];
 17 
 18 void add_edge(int u, int v, int w){
 19     edge[tot].w = w;
 20     edge[tot].to = v;
 21     edge[tot].next = head[u];
 22     head[u] = tot++;
 23     
 24     edge[tot].w = 0;
 25     edge[tot].to = u;
 26     edge[tot].next = head[v];
 27     head[v] = tot++;
 28 }
 29 
 30 struct Dinic{
 31     int level[N], S, T;
 32     void init(int _S, int _T){
 33         S = _S;
 34         T = _T;
 35         tot = 0;
 36         memset(head, -1, sizeof(head));
 37     }
 38     bool bfs(){
 39         queue<int> que;
 40         memset(level, -1, sizeof(level));
 41         level[S] = 0;
 42         que.push(S);
 43         while(!que.empty()){
 44             int u = que.front();
 45             que.pop();
 46             for(int i = head[u]; i != -1; i = edge[i].next){
 47                 int v = edge[i].to;
 48                 int w = edge[i].w;
 49                 if(level[v] == -1 && w > 0){
 50                     level[v] = level[u]+1;
 51                     que.push(v);
 52                 }
 53             }
 54         }
 55         return level[T] != -1;
 56     }
 57     int dfs(int u, int flow){
 58         if(u == T)return flow;
 59         int ans = 0, fw;
 60         for(int i = head[u]; i != -1; i = edge[i].next){
 61             int v = edge[i].to, w = edge[i].w;
 62             if(!w || level[v] != level[u]+1)
 63                   continue;
 64             fw = dfs(v, min(flow-ans, w));
 65             ans += fw;
 66             edge[i].w -= fw;
 67             edge[i^1].w += fw;
 68             if(ans == flow)return ans;
 69         }
 70         if(ans == 0)level[u] = -1;
 71         return ans;
 72     }
 73     int maxflow(){
 74         int flow = 0, f;
 75         while(bfs())
 76               while((f = dfs(S, INF)) > 0)
 77                 flow += f;
 78         return flow;    
 79     }
 80 }dinic;
 81 
 82 char str[N];
 83 int status[1<<12];//status[S]表示状态为S的人数，例如S=1010，表示可以在2号和4号星球上生存(从低位标号)
 84 
 85 int main()
 86 {
 87     std::ios::sync_with_stdio(false);    
 88     //freopen("inputM.txt", "r", stdin);
 89     int n, m;
 90     while(cin>>n>>m){
 91         int s = 0, t = n+m+1;
 92         dinic.init(s, t);
 93         int w;
 94         memset(status, 0, sizeof(status));
 95         for(int i = 1; i <= n; i++){
 96             int tmp = 0;
 97             for(int j = 1; j <= m; j++){
 98                 tmp <<= 1;
 99                 cin>>w;
100                 tmp |= w;
101             }
102             status[tmp]++;
103         }
104         for(int i = 1; i <= (1<<10); i++){
105             if(status[i]){
106                 add_edge(s, i, status[i]);
107                 for(int j = 1; j <= m; j++){
108                     if(i & (1<<(j-1)))
109                           add_edge(i, n+j, status[i]);
110                 }
111             }
112         }
113         int sum = 0;
114         for(int i = 1; i <= m; i++){
115             cin>>w;
116             sum += w;
117             add_edge(n+i, t, w);
118         }
119         if(sum < n){
120             cout<<"NO"<<endl;
121             continue;
122         }
123         if(dinic.maxflow() == n)cout<<"YES"<<endl;
124         else cout<<"NO"<<endl;
125     }    
126     return 0;
127 }

 

posted @ 2017-08-24 19:38  Penn000  阅读(265)  评论(0)  编辑  收藏  举报