HDU 3605 Escape (最大流+合并点)

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

注意n的范围是100000，m的范围是10，所以最多的情况是合并成2^10=1024个n

先用map合并点，然后源点到n的容量设为p(表示这种点的数量)，m到汇点的容量按题设

然后求最大流即可

PS:如果不合并点会TLE

  1 //#pragma comment(linker, "/STACK:102400000,102400000")
  2 #include<cstdio>
  3 #include<iostream>
  4 #include<cstring>
  5 #include<string>
  6 #include<cmath>
  7 #include<set>
  8 #include<list>
  9 #include<map>
 10 #include<iterator>
 11 #include<cstdlib>
 12 #include<vector>
 13 #include<queue>
 14 #include<stack>
 15 #include<algorithm>
 16 #include<functional>
 17 using namespace std;
 18 typedef long long LL;
 19 #define ROUND(x) round(x)
 20 #define FLOOR(x) floor(x)
 21 #define CEIL(x) ceil(x)
 22 const int maxn=1200;
 23 const int maxm=200010;
 24 const int inf=0x3f3f3f3f;
 25 const LL inf64=0x3f3f3f3f3f3f3f3fLL;
 26 const double INF=1e30;
 27 const double eps=1e-6;
 28 
 29 /**
 30 *最大流最小割：加各种优化的Dinic算法($O(V^2E)$)
 31 *输入：图(链式前向星),n(顶点个数,包含源汇),st(源),ed(汇)
 32 *输出：Dinic(NdFlow)(最大流),MinCut()(最小割)(需先求最大流)
 33 *打印路径方法：按反向边(i&1)的flow 找，或者按边的flow找
 34 */
 35 //const int maxn=0;
 36 //const int maxm=0;
 37 //const int inf=0x3f3f3f3f;
 38 struct Edge
 39 {
 40     int u,v;
 41     int cap,flow;
 42     int next;
 43 } edge[maxm];
 44 int head[maxn],edgeNum;//需初始化
 45 int n,m,d[maxn],cur[maxn];
 46 int st,ed;
 47 bool vis[maxn];
 48 void addSubEdge(int u,int v,int cap,int flow)
 49 {
 50     edge[edgeNum].u=u;
 51     edge[edgeNum].v=v;
 52     edge[edgeNum].cap=cap;
 53     edge[edgeNum].flow=flow;
 54     edge[edgeNum].next=head[u];
 55     head[u]=edgeNum++;
 56     cur[u]=head[u];
 57 }
 58 void addEdge(int u,int v,int cap)
 59 {
 60     addSubEdge(u,v,cap,0);
 61     addSubEdge(v,u,0,0);//注意加反向0 边
 62 }
 63 bool BFS()
 64 {
 65     queue<int> Q;
 66     memset(vis, 0, sizeof(vis));
 67     Q.push(st);
 68     d[st]=0;
 69     vis[st]=1;
 70     while (!Q.empty())
 71     {
 72         int u=Q.front();
 73         Q.pop();
 74         for(int i=head[u]; i!=-1; i=edge[i].next)
 75         {
 76             int v=edge[i].v;
 77             int w=edge[i].cap-edge[i].flow;
 78             if(w>0 && !vis[v])
 79             {
 80                 vis[v]=1;
 81                 Q.push(v);
 82                 d[v]=d[u]+1;
 83                 if(v==ed) return 1;
 84             }
 85         }
 86     }
 87     return false;
 88 }
 89 int Aug(int u, int a)
 90 {
 91     if (u==ed) return a;
 92     int aug=0, delta;
 93     for(int &i=cur[u]; i!=-1; i=edge[i].next)
 94     {
 95         int v=edge[i].v;
 96         int w=edge[i].cap-edge[i].flow;
 97         if (w>0 && d[v]==d[u]+1)
 98         {
 99             delta = Aug(v, min(a,w));
100             if (delta)
101             {
102                 edge[i].flow += delta;
103                 edge[i^1].flow -= delta;
104                 aug += delta;
105                 if (!(a-=delta)) break;
106             }
107         }
108     }
109     if (!aug) d[u]=-1;
110     return aug;
111 }
112 int Dinic(int NdFlow)
113 {
114     int flow=0;
115     while (BFS())
116     {
117         memcpy(cur,head,sizeof(int)*(n+1));
118         flow += Aug(st,inf);
119         /*如果超过指定流量就return 掉*/
120         if(NdFlow==inf) continue;
121         if(flow > NdFlow) break;
122     }
123     return flow;
124 }
125 
126 map<string,int> mp;
127 map<string,int>::iterator ite;
128 int N,M;
129 char strx[110];
130 int num[20];
131 void init()
132 {
133     memset(head,-1,sizeof(head));
134     edgeNum=0;
135     mp.clear();
136 }
137 void input()
138 {
139     if(scanf("%d%d",&N,&M)==EOF) exit(0);
140     getchar();
141     for(int i=0;i<N;i++)
142     {
143         gets(strx);
144         string str=strx;
145         mp[str]++;
146     }
147     for(int i=0;i<M;i++) scanf("%d",&num[i]);
148 }
149 void build()
150 {
151     n=mp.size()+M+2;
152     st=0,ed=mp.size()+M+1;
153     int i;
154     for(ite=mp.begin(),i=1;ite!=mp.end();ite++,i++)
155     {
156         addEdge(st,i,ite->second);
157         int m=1;
158         for(int j=0;j<ite->first.size();j++)
159         {
160             if(ite->first[j]=='1')
161             {
162                 addEdge(i,mp.size()+m,ite->second);
163             }
164             if(ite->first[j]=='0'||ite->first[j]=='1')
165             {
166                 m++;
167             }
168         }
169     }
170     for(i=mp.size()+1;i<=mp.size()+M;i++)
171     {
172         addEdge(i,ed,num[i-(mp.size()+1)]);
173     }
174 }
175 void solve()
176 {
177     build();
178     if(Dinic(inf)==N) puts("YES");
179     else puts("NO");
180 //    printf("%d\n",Dinic(inf));
181 }
182 void output()
183 {
184     //
185 }
186 int main()
187 {
188 //    std::ios_base::sync_with_stdio(false);
189 //    freopen("in.cpp","r",stdin);
190     while(1)
191     {
192         init();
193         input();
194         solve();
195         output();
196     }
197     return 0;
198 }
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posted @ 2013-10-12 15:03  xysmlx  阅读(344)  评论(0)  编辑  收藏  举报