P2057 [SHOI2007]善意的投票 / [JLOI2010]冠军调查
拿来练网络流的qwq
思路:如果i不同意,连边(i,t,1),否则连边(s,i,1).好朋友x,y间连边(x,y,1)(y,x,1),最小割即为答案。每割一条边表示与他的意愿不相符。
#include<bits/stdc++.h>
using namespace std;
const int N=305;
const int inf=1000000007;
int head[N],num_edge=1;
struct edge{
int nxt,val,to;
}e[N*N*2];
inline void add(int from,int to,int val)
{
++num_edge;
e[num_edge].nxt=head[from];
e[num_edge].to=to;
e[num_edge].val=val;
head[from]=num_edge;
}
int n,m,s,t;
int maxflow,dep[N],cur[N];
bool inq[N];
bool bfs()
{
for(int i=1;i<=t;++i)
inq[i]=0,dep[i]=inf,cur[i]=head[i];
dep[s]=1;
queue<int>q;
q.push(s);
while(!q.empty())
{
int u=q.front();q.pop();
inq[u]=0;
for(int i=head[u];i;i=e[i].nxt)
{
int v=e[i].to;
if(dep[v]>dep[u]+1&&e[i].val)
{
dep[v]=dep[u]+1;
if(!inq[v])q.push(v),inq[v]=1;
}
}
}
return dep[t]!=inf;
}
int dfs(int u,int flow)
{
if(u==t)
{
maxflow+=flow;
return flow;
}
int used=0,rlow;
for(int i=cur[u];i;i=e[i].nxt)
{
cur[u]=i;
int v=e[i].to;
if(dep[v]==dep[u]+1&&e[i].val)
{
rlow=dfs(v,min(flow-used,e[i].val));
if(rlow)
{
used+=rlow;
e[i].val-=rlow;
e[i^1].val+=rlow;
if(used==flow)break;
}
}
}
return used;
}
void dinic(){while(bfs())dfs(s,inf);}
int main()
{
scanf("%d%d",&n,&m);s=n+1;t=n+2;
for(int i=1,x;i<=n;++i)
{
scanf("%d",&x);
if(x==0)add(i,t,1),add(t,i,0);
if(x==1)add(s,i,1),add(i,s,0);
}
for(int i=1,x,y;i<=m;++i)
{
scanf("%d%d",&x,&y);
add(x,y,1);add(y,x,0);
add(y,x,1);add(x,y,0);
}
dinic();
printf("%d\n",maxflow);
return 0;
}