把人分成两个集合,一个赞成睡觉,一个反对睡觉。好朋友连一条容量为1的双向边,s向赞成睡觉的连边,反对睡觉的向t连边。
那么这个图的一个割就对应着一个方案。如果割掉s和v的边,就代表v投意见与它自己相反的票,t和v的边同理。割掉u和v的边,就代表了这对好朋友之间意见不同。
这样求出一个割之后,好朋友之间意见不同的边都被割去了。
求出此图的最小割即为答案。
# include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # include <vector> # include <queue> # include <stack> # include <map> # include <set> # include <cmath> # include <algorithm> using namespace std; # define lowbit(x) ((x)&(-x)) # define pi acos(-1.0) # define eps 1e-9 # define MOD 1024523 # define INF 1000000000 # define mem(a,b) memset(a,b,sizeof(a)) # define FOR(i,a,n) for(int i=a; i<=n; ++i) # define FO(i,a,n) for(int i=a; i<n; ++i) # define bug puts("H"); # define lch p<<1,l,mid # define rch p<<1|1,mid+1,r # define mp make_pair # define pb push_back typedef pair<int,int> PII; typedef vector<int> VI; # pragma comment(linker, "/STACK:1024000000,1024000000") typedef long long LL; int Scan() { int x=0,f=1;char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();} return x*f; } void Out(int a) { if(a<0) {putchar(‘-‘); a=-a;} if(a>=10) Out(a/10); putchar(a%10+‘0‘); } const int N=305; //Code begin... struct Edge{int p, next, w;}edge[200005]; int head[N], cnt=2, s, t, vis[N]; queue<int>Q; void add_edge(int u, int v, int w){ edge[cnt].p=v; edge[cnt].w=w; edge[cnt].next=head[u]; head[u]=cnt++; edge[cnt].p=u; edge[cnt].w=0; edge[cnt].next=head[v]; head[v]=cnt++; } int bfs(){ int i, v; mem(vis,-1); vis[s]=0; Q.push(s); while (!Q.empty()) { v=Q.front(); Q.pop(); for (i=head[v]; i; i=edge[i].next) { if (edge[i].w>0 && vis[edge[i].p]==-1) { vis[edge[i].p]=vis[v] + 1; Q.push(edge[i].p); } } } return vis[t]!=-1; } int dfs(int x, int low){ int i, a, temp=low; if (x==t) return low; for (i=head[x]; i; i=edge[i].next) { if (edge[i].w>0 && vis[edge[i].p]==vis[x]+1){ a=dfs(edge[i].p,min(edge[i].w,temp)); temp-=a; edge[i].w-=a; edge[i^1].w += a; if (temp==0) break; } } if (temp==low) vis[x]=-1; return low-temp; } int main () { int n, m, flag, u, v; scanf("%d%d",&n,&m); s=0; t=n+1; FOR(i,1,n) { scanf("%d",&flag); if (flag) add_edge(s,i,1); else add_edge(i,t,1); } FOR(i,1,m) scanf("%d%d",&u,&v), add_edge(u,v,1), add_edge(v,u,1); int sum=0, tmp; while (bfs()) while (tmp=dfs(s,INF)) sum+=tmp; printf("%d\n",sum); return 0; }
时间: 2024-10-05 11:43:20