【BZOJ】【2768】【JLOI2010】冠军调查

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　　我不会告诉你这题跟 BZOJ 1934 是一模一样的……包括数据范围……
  1 /**************************************************************
  2     Problem: 2768
  3     User: Tunix
  4     Language: C++
  5     Result: Accepted
  6     Time:32 ms
  7     Memory:4408 kb
  8 ****************************************************************/
  9  
 10 //BZOJ 2768
 11 #include<vector>
 12 #include<cstdio>
 13 #include<cstring>
 14 #include<cstdlib>
 15 #include<iostream>
 16 #include<algorithm>
 17 #define rep(i,n) for(int i=0;i<n;++i)
 18 #define F(i,j,n) for(int i=j;i<=n;++i)
 19 #define D(i,j,n) for(int i=j;i>=n;--i)
 20 #define pb push_back
 21 using namespace std;
 22 inline int getint(){
 23     int v=0,sign=1; char ch=getchar();
 24     while(ch<'0'||ch>'9'){ if (ch=='-') sign=-1; ch=getchar();}
 25     while(ch>='0'&&ch<='9'){ v=v*10+ch-'0'; ch=getchar();}
 26     return v*sign;
 27 }
 28 const int N=320,M=200000,INF=~0u>>2;
 29 typedef long long LL;
 30 /******************tamplate*********************/
 31 int n,m;
 32 struct edge{
 33     int from,to,v;
 34 };
 35 inline int pack(int i,int j){return (i-1)*m+j;}
 36 struct Net{
 37     edge E[M];
 38     int head[N],next[M],cnt;
 39     void add(int x,int y,int v){
 40         E[++cnt]=(edge){x,y,v};
 41         next[cnt]=head[x]; head[x]=cnt;
 42         E[++cnt]=(edge){y,x,0};
 43         next[cnt]=head[y]; head[y]=cnt;
 44     }
 45     int s,t,cur[N],d[N],Q[N];
 46     void init(){
 47         n=getint();m=getint();
 48         cnt=1;
 49         s=0; t=n+1;
 50         F(i,1,n)
 51             if (getint()) add(s,i,1);
 52             else add(i,t,1);
 53         int x,y;
 54         F(i,1,m){
 55             x=getint(); y=getint();
 56             add(x,y,1);
 57             add(y,x,1);
 58         }
 59     }
 60     bool mklevel(){
 61         memset(d,-1,sizeof d);
 62         d[s]=0;
 63         int l=0,r=-1;
 64         Q[++r]=s;
 65         while(l<=r){
 66             int x=Q[l++];
 67             for(int i=head[x];i;i=next[i])
 68                 if (d[E[i].to]==-1 && E[i].v){
 69                     d[E[i].to]=d[x]+1;
 70                     Q[++r]=E[i].to;
 71                 }
 72         }
 73         return d[t]!=-1;
 74     }
 75     int dfs(int x,int a){
 76         if (x==t||a==0) return a;
 77         int flow=0;
 78         for(int &i=cur[x];i && flow<a;i=next[i])
 79             if (d[E[i].to]==d[x]+1 && E[i].v){
 80                 int f=dfs(E[i].to,min(a-flow,E[i].v));
 81                 E[i].v-=f;
 82                 E[i^1].v+=f;
 83                 flow+=f;
 84             }
 85         if (!flow) d[x]=-1;
 86         return flow;
 87     }
 88     int Dinic(){
 89         int flow=0;
 90         while(mklevel()){
 91             F(i,s,t) cur[i]=head[i];
 92             flow+=dfs(s,INF);
 93         }
 94         return flow;
 95     }
 96 }G1;
 97 int main(){
 98 #ifndef ONLINE_JUDGE
 99     freopen("2768.in","r",stdin);
100     freopen("2768.out","w",stdout);
101 #endif
102     G1.init();
103     printf("%d
",G1.Dinic());
104     return 0;
105 }
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