2011 Multi-University Training Contest 1 - Host by HNU
A.A + B problem(待填坑)
B.Cat VS Dog(二分图匹配)
喜欢cat和喜欢dog的人构成了二分图，如果两个人有冲突则连一条边，则问题转化为二分图最大点独立集问题。ans=n-最大匹配。
# 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-3 # define MOD 1000000007 # 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 res=0, flag=0; char ch; if((ch=getchar())=='-') flag=1; else if(ch>='0'&&ch<='9') res=ch-'0'; while((ch=getchar())>='0'&&ch<='9') res=res*10+(ch-'0'); return flag?-res:res; } void Out(int a) { if(a<0) {putchar('-'); a=-a;} if(a>=10) Out(a/10); putchar(a%10+'0'); } const int N=505; //Code begin... struct Child{string like, dis;}cat[N], dog[N]; int G[N][N], link[N], used[N], cat_num, dog_num; int dfs(int k) { FO(i,0,dog_num) if(!used[i] && G[k][i]) { used[i]=1; if(link[i]==-1 || dfs(link[i])) {link[i]=k; return 1;} } return 0; } int maxMatch() { int cnt=0; FO(i,0,cat_num) { mem(used,0); if(dfs(i)) ++cnt; } return cnt; } int main() { int n, m, p; string a, b; while(cin>>n>>m>>p) { mem(G,0); mem(link,-1); cat_num=dog_num=0; while(p--) { cin>>a>>b; if(a[0] == 'C') cat[cat_num].like=a, cat[cat_num].dis=b, cat_num++; else dog[dog_num].like=a, dog[dog_num].dis=b, dog_num++; } FO(i,0,cat_num) FO(j,0,dog_num) if(cat[i].like==dog[j].dis || cat[i].dis==dog[j].like) G[i][j]=1; cout<<cat_num+dog_num-maxMatch()<<endl; } return 0; }
C.Checkers(二分+LCA+辗转相除法)
考虑点对(a,b,c)可以转换的点对。容易发现这其实是一个无限二叉树。
如果左边的棋子或者右边的棋子往中间跳，就对应着父亲边往上的过程。中间的棋子往左跳，对应着左孩子边。中间的往右跳，对应着右孩子边。
那么我们判断解的存在性就相当于找到他们是不是同一个根。判断最小的解就相当于在树上找到LCA。
可以通过辗转相除法来优化向上爬的操作。由于不能倍增找LCA，我们二分向上爬的深度验证即可。
# 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-3 # define MOD 100000007 # 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 res=0, flag=0; char ch; if((ch=getchar())=='-') flag=1; else if(ch>='0'&&ch<='9') res=ch-'0'; while((ch=getchar())>='0'&&ch<='9') res=res*10+(ch-'0'); return flag?-res:res; } void Out(int a) { if(a<0) {putchar('-'); a=-a;} if(a>=10) Out(a/10); putchar(a%10+'0'); } const int N=505; //Code begin... LL a[3], b[3], root[2], dep[2], dis[2], temp1[3], temp2[3]; void get_root(LL x) { root[x]=(x?b[1]:a[1]); dep[x]=0; LL val1, val2, t; if (x) val1=b[1]-b[0], val2=b[2]-b[1]; else val1=a[1]-a[0], val2=a[2]-a[1]; while (val1!=val2) { if (val1<val2) { if (val2%val1) t=val2/val1; else t=val2/val1-1; root[x]+=t*val1; dep[x]+=t; val2-=t*val1; } else { if (val1%val2) t=val1/val2; else t=val1/val2-1; root[x]-=t*val2; dep[x]+=t; val1-=t*val2; } } dis[x]=val1; } void go_up(LL (&a)[3], LL dep) { while (dep) { LL val1=a[1]-a[0], val2=a[2]-a[1], t; if (val1>val2) { if (val1%val2) t=val1/val2; else t=val1/val2-1; if (dep>=t) a[1]-=t*val2, a[2]-=t*val2, dep-=t; else a[1]-=dep*val2, a[2]-=dep*val2, dep=0; } else { if (val2%val1) t=val2/val1; else t=val2/val1-1; if (dep>=t) a[1]+=t*val1, a[0]+=t*val1, dep-=t; else a[1]+=dep*val1, a[0]+=dep*val1, dep=0; } } } int main () { while (~scanf("%I64d%I64d%I64d%I64d%I64d%I64d",&a[0],&a[1],&a[2],&b[0],&b[1],&b[2])) { sort(a,a+3); sort(b,b+3); get_root(0); get_root(1); if (root[0]!=root[1] || dis[0]!=dis[1]) {puts("NO"); continue;} LL ans=0; if (dep[0]>dep[1]) { LL t=dep[0]-dep[1]; dep[0]=dep[1]; go_up(a,t); ans+=t; } else if (dep[0]<dep[1]) { LL t=dep[1]-dep[0]; dep[1]=dep[0]; go_up(b,t); ans+=t; } if (a[0]==b[0]&&a[1]==b[1]&&a[2]==b[2]) {printf("YES\n%I64d\n",ans); continue;} LL l=0, r=dep[0]+1, mid; while (l<r) { mid=(l+r)>>1; temp1[0]=a[0]; temp1[1]=a[1]; temp1[2]=a[2]; temp2[0]=b[0]; temp2[1]=b[1]; temp2[2]=b[2]; go_up(temp1,mid); go_up(temp2,mid); if (temp1[0]==temp2[0]&&temp1[1]==temp2[1]&&temp1[2]==temp2[2]) r=mid; else l=mid+1; } printf("YES\n%I64d\n",ans+2*r); } return 0; }
D.DICS(DP)
这是字符串编辑距离的拓展，多了一个改变后缀操作。dp数组多开一维记录改变的后缀类型。直接记搜即可。
# 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-3 # define MOD 100000007 # 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 res=0, flag=0; char ch; if((ch=getchar())=='-') flag=1; else if(ch>='0'&&ch<='9') res=ch-'0'; while((ch=getchar())>='0'&&ch<='9') res=res*10+(ch-'0'); return flag?-res:res; } void Out(int a) { if(a<0) {putchar('-'); a=-a;} if(a>=10) Out(a/10); putchar(a%10+'0'); } const int N=505; //Code begin... int dp[N][N][55], len1, len2; char s1[N], s2[N]; int dfs(int x, int y, int z) { if (~dp[x][y][z]) return dp[x][y][z]; if (x==len1&&y==len2) return 0; if (x==len1) return len2-y; if (y==len2) return len1-x; int ans=min(dfs(x,y+1,z)+1,dfs(x+1,y,z)+1); if (z==0) ans=min(ans,dfs(x+1,y+1,z)+(s1[x]!=s2[y])); else ans=min(ans,dfs(x+1,y+1,z)+(z!=s2[y])); if ((z==0&&s1[x]!=s2[y]) || (z&&z!=s2[y])) ans=min(ans,dfs(x+1,y+1,s2[y])+1); return dp[x][y][z]=ans; } int main () { while (1) { scanf("%s",s1); if (s1[0]=='#') break; scanf("%s",s2); len1=strlen(s1); len2=strlen(s2); FO(i,0,len1) { if (s1[i]<='z'&&s1[i]>='a') s1[i]=s1[i]-'a'+1; else s1[i]=s1[i]-'A'+27; } FO(i,0,len2) { if (s2[i]<='z'&&s2[i]>='a') s2[i]=s2[i]-'a'+1; else s2[i]=s2[i]-'A'+27; } mem(dp,-1); printf("%d\n",dfs(0,0,0)); } return 0; }
E.Earth Hour(水题)
先抽象成图，然后问题转化为求图中3个点的生成树的最短路。数据范围很小，直接从这3个点bfs一次。因为这连接3个点的生成树一定是从图中一个点出发可以直接到达的。
枚举这个点更新答案即可。
# 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-3 # define MOD 100000007 # 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 res=0, flag=0; char ch; if((ch=getchar())=='-') flag=1; else if(ch>='0'&&ch<='9') res=ch-'0'; while((ch=getchar())>='0'&&ch<='9') res=res*10+(ch-'0'); return flag?-res:res; } void Out(int a) { if(a<0) {putchar('-'); a=-a;} if(a>=10) Out(a/10); putchar(a%10+'0'); } const int N=205; //Code begin... struct Edge{int p, next;}edge[N*N]; struct Cir{int x, y, r;}cir[N]; int head[N], cnt, fa[N], vis[N]; queue<int>Q; int find(int x) { int s, temp; for (s=x; fa[s]>=0; s=fa[s]) ; while (s!=x) temp=fa[x], fa[x]=s, x=temp; return s; } void union_set(int x, int y) { int temp=fa[x]+fa[y]; if (fa[x]>fa[y]) fa[x]=y, fa[y]=temp; else fa[y]=x, fa[x]=temp; } void add_edge(int u, int v){edge[cnt].p=v; edge[cnt].next=head[u]; head[u]=cnt++;} void init(){cnt=1; mem(head,0); mem(fa,-1);} int dis(int x, int y){return (cir[x].x-cir[y].x)*(cir[x].x-cir[y].x)+(cir[x].y-cir[y].y)*(cir[x].y-cir[y].y);} int main () { int T, n, u, v, w; scanf("%d",&T); while (T--) { int ans=0; init(); scanf("%d",&n); FOR(i,1,n) scanf("%d%d%d",&cir[i].x,&cir[i].y,&cir[i].r); FOR(i,1,n) FOR(j,i+1,n) if (dis(i,j)<=(cir[i].r+cir[j].r)*(cir[i].r+cir[j].r)) { add_edge(i,j), add_edge(j,i); u=find(i), v=find(j); if (u!=v) union_set(u,v); } u=find(1); v=find(2); w=find(3); if (u!=v||u!=w||v!=w) {puts("-1"); continue;} FOR(i,1,n) { mem(vis,-1); vis[i]=0; while (!Q.empty()) Q.pop(); Q.push(i); while (!Q.empty()) { int w=Q.front(); Q.pop(); for (int v=head[w]; v; v=edge[v].next) { int u=edge[v].p; if (vis[u]!=-1) continue; vis[u]=vis[w]+1; Q.push(u); } } if (vis[1]!=-1&&vis[2]!=-1&&vis[3]!=-1) ans=max(ans,n-1-vis[1]-vis[2]-vis[3]); } printf("%d\n",ans); } }
F.YY's new problem(线段树维护字符串hash)
问题转化为 2*a[j]=a[i]+a[k].对于每一个a[j], 只需要验证点对(a[i],a[k])是否出现在a[j]的两侧。复杂度是O(n^2).
如果发现对于a[j],点对是连续的,比如(a[j]-1,a[j]+1),(a[j]-2,a[j]+2),(a[j]-k,a[j]+k).
从前往后扫描数组，如果扫描到a[m],就将vis[a[m]]置1，扫描到a[j]时，只需要验证每个点对的vis值是否是相等的。这样复杂度还是O(n^2).
实际上点对的分布在数轴上是关于点a[j]对称的，那么我们只需要检查关于vis值字符串a[j]-k...a[j]-2,a[j]-1和字符串a[j]+k...a[j]+2,a[j]+1.是否相等就行了。
于是我们需要一个能够快速实现更新一个点的hash，查询一个区间的hash值的数据结构，hash由于具有区间合并性。我们可以用线段树来维护区间的hash值。
复杂度是O(nlogn).
# 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-3 # define MOD 100000007 # 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 res=0, flag=0; char ch; if((ch=getchar())=='-') flag=1; else if(ch>='0'&&ch<='9') res=ch-'0'; while((ch=getchar())>='0'&&ch<='9') res=res*10+(ch-'0'); return flag?-res:res; } void Out(int a) { if(a<0) {putchar('-'); a=-a;} if(a>=10) Out(a/10); putchar(a%10+'0'); } const int N=10005; //Code begin... int a[N]; LL seg1[N<<2], seg2[N<<2], _hash[N]; LL mult_mod(LL a, LL b, LL c) { a%=c; b%=c; LL ret=0, tmp=a; while (b) { if (b&1) { ret+=tmp; if (ret>c) ret-=c; } tmp<<=1; if (tmp>c) tmp-=c; b>>=1; } return ret; } LL pow_mod(LL a, LL n, LL mod) { LL ret=1, temp=a%mod; while (n) { if (n&1) ret=mult_mod(ret,temp,mod); temp=mult_mod(temp,temp,mod); n>>=1; } return ret; } void init(int p, int l, int r) { if (l<r) { int mid=(l+r)>>1; init(lch); init(rch); seg1[p]=(seg1[p<<1]*_hash[(r-l+1)>>1]+seg1[p<<1|1])%MOD; seg2[p]=(seg2[p<<1|1]*_hash[(r-l+2)>>1]+seg2[p<<1])%MOD; return ; } seg1[p]=seg2[p]=1; } void insert(int p, int l, int r, int L) { if (L>r||L<l) return ; if (L==l&&L==r) seg1[p]=seg2[p]=2; else { int mid=(l+r)>>1; insert(lch,L); insert(rch,L); seg1[p]=(seg1[p<<1]*_hash[(r-l+1)>>1]+seg1[p<<1|1])%MOD; seg2[p]=(seg2[p<<1|1]*_hash[(r-l+2)>>1]+seg2[p<<1])%MOD; } } LL query1(int p, int l, int r, int L, int R) { if (L>r||R<l) return 0; if (L<=l&&R>=r) return seg1[p]*_hash[R-r]%MOD; int mid=(l+r)>>1; return (query1(lch,L,R)+query1(rch,L,R))%MOD; } LL query2(int p, int l, int r, int L, int R) { if (L>r||R<l) return 0; if (L<=l&&R>=r) return seg2[p]*_hash[l-L]%MOD; int mid=(l+r)>>1; return (query2(lch,L,R)+query2(rch,L,R))%MOD; } int main () { int n, T, flag; _hash[0]=1; FOR(i,1,10000) _hash[i]=_hash[i-1]*131%MOD; scanf("%d",&T); while (T--) { flag=0; scanf("%d",&n); FOR(i,1,n) scanf("%d",a+i); init(1,1,n); FOR(i,1,n) { if (a[i]!=1 && a[i]!=n) { int t=min(n-a[i],a[i]-1); //printf("%d %lld %lld\n",i,query1(1,1,n,a[i]-t,a[i]-1),query2(1,1,n,a[i]+1,a[i]+t)); if (query1(1,1,n,a[i]-t,a[i]-1)!=query2(1,1,n,a[i]+1,a[i]+t)) {flag=1; break;} } insert(1,1,n,a[i]); } puts(flag?"Y":"N"); } return 0; }
G.Where am I(计算几何)
把小圆缩成点，大圆缩小，可以发现路径只会形成一个三角形，计算出这个三角形即可。
#include <stdio.h> #include <string.h> #include <stdlib.h> #include <math.h> const double eps = 1e-6; const double pi = acos(-1.0); typedef struct { double x,y; }TPoint; typedef struct { double x,y,r; }TCircle; TCircle ball,sball; TPoint vec,convert; double T; int flag; inline double squ(double x) { return x*x; } double dot(TPoint a,TPoint b,TPoint c) { return (b.x-a.x)*(c.x-a.x)+(b.y-a.y)*(c.y-a.y); } int cross(TPoint a,TPoint b,TPoint c) { double ret = (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x); if(ret>eps) return 1; return -1; } void work() { convert.x = ball.x; convert.y = ball.y; sball.x -= convert.x; sball.y -= convert.y; ball.x =0; ball.y = 0; ball.r -= sball.r; TPoint me = {sball.x,sball.y}; if(fabs(vec.x) < eps && fabs(vec.y) < eps) {/*vec is zero*/ me.x += convert.x; me.y += convert.y; printf("%.1f %.1f\n",me.x,me.y); return; } if(fabs(ball.r)<eps) {/*sball.r == ball.r*/ me.x += convert.x; me.y += convert.y; printf("%.1f %.1f\n",me.x,me.y); return; } if(squ(me.x+T*vec.x)+squ(me.y+T*vec.y)+eps<squ(ball.r)) { me.x = me.x+T*vec.x; me.y = me.y+T*vec.y; me.x += convert.x; me.y += convert.y; printf("%.1f %.1f\n",me.x,me.y); return ; } double k=vec.y/vec.x; double b=me.y-k*me.x; double x1; //printf("k:%f b:%f %f\n",k,b,ball.r); if(vec.x<0) { x1=(-k*b-sqrt(squ(k*ball.r)+squ(ball.r)-squ(b)))/(squ(k)+1); } else { x1=(-k*b+sqrt(squ(k*ball.r)+squ(ball.r)-squ(b)))/(squ(k)+1.0); } /*fjjjjj*/ double y1=k*x1+b; /** ** x1 && y1 **/ // printf("(x1,y1)%f %f\n",x1,y1); TPoint tmp1 = {x1,y1}; TPoint tmp2 = {0,0}; double u = dot(tmp1,tmp2,me); flag = cross(tmp1,tmp2,me); /** flag **/ //printf("flag:%d\n",flag); u = u/(sqrt(squ(x1-me.x)+squ(y1-me.y))*ball.r); double l=(u)*ball.r; l *= 2; u = acos(u); /** l **/ //printf("l:%f\n",l); double t1 = l/sqrt(squ(vec.x)+squ(vec.y)); __int64 time = (T-(x1-me.x)/vec.x)/t1; /** time **/ //printf("time:%I64d\n",time); T = T-(x1-me.x)/vec.x-time*t1; /** lefttime **/ // printf("lefttime:%f\n",T); /*上面以对 很可能是角度算错了 下面的要仔细验证下!!!*/ double u2=atan2(y1,x1); double u3 = (time+1)*(pi-2*u)*flag; u2 += time*(pi-2*u)*flag; double x2 = ball.r*cos(u2); double y2 = ball.r*sin(u2); //printf("%f %f(x2,y2)\n",x2,y2); TPoint vec2; vec2.x = vec.x*cos(u3)-vec.y*sin(u3); vec2.y = vec.x*sin(u3)+vec.y*cos(u3); me.x = x2+vec2.x*T+convert.x; me.y = y2+vec2.y*T+convert.y; printf("%.1f %.1f\n",me.x,me.y); return; } int main(void) { #ifndef ONLINE_JUDGE freopen("G.in","r",stdin); freopen("G.out","w",stdout); #endif int t; scanf("%d",&t); while(t--) { scanf("%lf%lf%lf",&ball.x,&ball.y,&ball.r); scanf("%lf%lf%lf",&sball.x,&sball.y,&sball.r); scanf("%lf%lf%lf",&vec.x,&vec.y,&T); work(); } return 0; }
H.R(N)(水题)
打出平方表二分搜一下即可。
# 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-3 # define MOD 1000000007 # 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 res=0, flag=0; char ch; if((ch=getchar())=='-') flag=1; else if(ch>='0'&&ch<='9') res=ch-'0'; while((ch=getchar())>='0'&&ch<='9') res=res*10+(ch-'0'); return flag?-res:res; } void Out(int a) { if(a<0) {putchar('-'); a=-a;} if(a>=10) Out(a/10); putchar(a%10+'0'); } const int N=100005; //Code begin... int a[N], P; bool bin_sea(int x) { int l=1, r=P+1, mid; while (l<r) { mid=(l+r)>>1; if (l==mid) break; if (a[mid]>x) r=mid; else l=mid; } return a[l]==x; } int main () { int n; P=(int)sqrt(1e9+0.5); FOR(i,1,P) a[i]=i*i; while (~scanf("%d",&n)) { if (n==0) {puts("1"); continue;} LL ans=0; if (bin_sea(n)) ans+=4; FOR(i,1,n) { if (2*i*i>n) break; else if (2*i*i==n) ans+=4; else if (bin_sea(n-i*i)) ans+=8; } printf("%lld\n",ans); } return 0; }
I.Equivalent Sets(强连通分量)
易知道当且仅当一个图是强连通图时题意成立。先把题中给出的图强连通分量缩点，然后变成了一个DAG，问题转化为求DAG最少加几条边变成强连通图。
有结论max(源，汇)。(具体还不会证)
# 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-3 # define MOD 100000007 # 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 res=0, flag=0; char ch; if((ch=getchar())=='-') flag=1; else if(ch>='0'&&ch<='9') res=ch-'0'; while((ch=getchar())>='0'&&ch<='9') res=res*10+(ch-'0'); return flag?-res:res; } void Out(int a) { if(a<0) {putchar('-'); a=-a;} if(a>=10) Out(a/10); putchar(a%10+'0'); } const int N=20005; //Code begin... struct Edge{int p, next;}edge[N<<2]; int head[N], cnt; int Low[N], DFN[N], Stack[N], Belong[N], num[N]; int Index, top, scc, a[N<<2][2], chu[N], ru[N]; bool Instack[N]; void add_edge(int u, int v){edge[cnt].p=v; edge[cnt].next=head[u]; head[u]=cnt++;} void Tarjan(int u) { int v; Low[u]=DFN[u]=++Index; Stack[top++]=u; Instack[u]=true; for (int i=head[u]; i; i=edge[i].next) { v=edge[i].p; if (!DFN[v]) { Tarjan(v); if (Low[u]>Low[v]) Low[u]=Low[v]; } else if (Instack[v]&&Low[u]>DFN[v]) Low[u]=DFN[v]; } if (Low[u]==DFN[u]) { scc++; do{ v=Stack[--top]; Instack[v]=false; Belong[v]=scc; num[scc]++; }while (v!=u); } } void solve(int n) { mem(DFN,0); mem(Instack,0); mem(num,0); Index=scc=top=0; FOR(i,1,n) if (!DFN[i]) Tarjan(i); } void init(){mem(head,0); mem(chu,0); mem(ru,0); cnt=1;} int main () { int n, m, u, v; while (~scanf("%d%d",&n,&m)) { init(); FOR(i,1,m) scanf("%d%d",&u,&v), add_edge(u,v), a[i][0]=u, a[i][1]=v; solve(n); if (scc==1) {puts("0"); continue;} FOR(i,1,m) { if (Belong[a[i][0]]==Belong[a[i][1]]) continue; ++chu[Belong[a[i][0]]]; ++ru[Belong[a[i][1]]]; } int ans1=0, ans2=0; FOR(i,1,scc) { if (chu[i]==0) ans1++; if (ru[i]==0) ans2++; } printf("%d\n",max(ans1,ans2)); } return 0; }
