Description
Input
Output
最多可选多少景点
Sample Input
7 6
1 2
2 3
5 4
4 3
3 6
6 7
1 2
2 3
5 4
4 3
3 6
6 7
Sample Output
2
HINT
这题是结论题
答案=最长反链=最小路径覆盖=n-二分图最大匹配
先floyd处理出两点之间的联通性,然后拆点,如果A能到B则A向B'连边
#include<cstdio> #include<iostream> #include<cstring> #define LL long long #define inf 0x7ffffff #define S 0 #define T 2*n+1 using namespace std; inline LL read() { LL 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; } struct edge{int to,next,v;}e[100010]; int n,m,cnt=1,ans; bool go[1010][1010]; int head[100010]; int q[100010]; int h[100010]; inline void ins(int u,int v,int w) { e[++cnt].to=v; e[cnt].v=w; e[cnt].next=head[u]; head[u]=cnt; } inline void insert(int u,int v,int w) { ins(u,v,w); ins(v,u,0); } inline bool bfs() { memset(h,-1,sizeof(h)); q[1]=S;h[S]=0;int t=0,w=1; while (t<w) { int now=q[++t]; for (int i=head[now];i;i=e[i].next) if (e[i].v&&h[e[i].to]==-1) { q[++w]=e[i].to; h[e[i].to]=h[now]+1; } } if (h[T]==-1)return 0; return 1; } inline int dfs(int x,int f) { if (x==T||!f)return f; int w,used=0; for (int i=head[x];i;i=e[i].next) if (e[i].v&&h[e[i].to]==h[x]+1) { w=dfs(e[i].to,min(f-used,e[i].v)); e[i].v-=w; e[i^1].v+=w; used+=w; if (used==f)return f; } if (!used)h[x]=-1; return used; } inline void dinic() {while (bfs())ans+=dfs(S,inf);} int main() { n=read();m=read(); for (int i=1;i<=m;i++) { int x=read(),y=read(); go[x][y]=1; } for (int k=1;k<=n;k++) for (int i=1;i<=n;i++) for (int j=1;j<=n;j++) if (go[i][k]&&go[k][j])go[i][j]=1; for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) if (go[i][j])insert(i,j+n,1); for (int i=1;i<=n;i++) insert(S,i,1),insert(i+n,T,1); dinic(); printf("%d ",n-ans); }