本以为会是三道小强与阿米巴，结果打开题目一看发现了这个：
T1:
恩先写着一道
#include<cstdio> #include<cstring> #include<cctype> #include<algorithm> #define rep(s,t) for(int i=s;i<=t;i++) using namespace std; inline int read() { char ch=getchar();int x=0,f=1; for(;!isdigit(ch);ch=getchar()) if(ch=='-') f=-1; for(;isdigit(ch);ch=getchar()) x=x*10+ch-'0'; return x*f; } const int maxn=100010; const int maxnode=2000010; int A[maxn],B[maxn],root[maxn]; int ls[maxnode],rs[maxnode],s[maxnode],ToT; void build(int& y,int x,int l,int r,int pos) { s[y=++ToT]=s[x]+1;if(l==r) return; int mid=l+r>>1;ls[y]=ls[x];rs[y]=rs[x]; if(pos<=mid) build(ls[y],ls[x],l,mid,pos); else build(rs[y],rs[x],mid+1,r,pos); } int query(int y,int x,int l,int r,int k) { if(l==r) return l; int k2=s[ls[y]]-s[ls[x]],mid=l+r>>1; if(k<=k2) return query(ls[y],ls[x],l,mid,k); return query(rs[y],rs[x],mid+1,r,k-k2); } int main() { freopen("kth.in","r",stdin); freopen("kth.out","w",stdout); int n=read(),q=read(); rep(1,n) A[i]=B[i]=read(); sort(B+1,B+n+1); rep(1,n) build(root[i],root[i-1],1,n,lower_bound(B+1,B+n+1,A[i])-B); rep(1,q) { int l=read(),r=read(),k=read(); printf("%d ",B[query(root[r],root[l-1],1,n,r-l-k+2)]); } return 0; }
T2:
无脑爆搜+调整顺序，只搞出来5个点。
#include<cstdio> #include<cstring> #include<cctype> #include<algorithm> #define rep(i,s,t) for(int i=s;i<=t;i++) using namespace std; inline int read() { char ch=getchar();int x=0,f=1; for(;!isdigit(ch);ch=getchar()) if(ch=='-') f=-1; for(;isdigit(ch);ch=getchar()) x=x*10+ch-'0'; return x*f; } int cnt; char s[20][20]; int rr[423],ry[324]; int id(int x,int y) {return (((y-1)/4)*4+(x-1)/4)+1;} int r[20][20],c[20][20],sq[20][20],num[20][20],use[17]; struct Point { int x,y,s; bool operator < (const Point& ths) const { return rr[x]+ry[num[x][y]]>rr[ths.x]+ry[num[x][y]];//s>ths.s; } }A[410]; void dfs(int cur) { if(cur==cnt+1) { rep(i,1,16) { rep(j,1,16) putchar(s[i][j]); puts(""); } exit(0); } else { int x=A[cur].x,y=A[cur].y,z=num[x][y]; for(int d=1;d<=16;d++) if(!r[x][d]&&!c[y][d]&&!sq[z][d]){ r[x][d]=c[y][d]=sq[z][d]=1;use[d]++; s[x][y]=(d<=10?d-1+'0':d-11+'A');dfs(cur+1); r[x][d]=c[y][d]=sq[z][d]=0;use[d]--; } } } int sc[1010]; int main() { srand(time(0)); rr[6]=17;rr[7]=876;rr[10]=16;rr[12]=12;ry[10]=38;ry[6]=34; freopen("sixteen7.in","r",stdin); freopen("sixteen7.out","w",stdout); sc['0']=1;sc['1']=2;sc['2']=3;sc['3']=4; sc['4']=5;sc['5']=6;sc['6']=7;sc['7']=8; sc['8']=9;sc['9']=10;sc['A']=11;sc['B']=12; sc['C']=13;sc['D']=14;sc['E']=15;sc['F']=16; rep(i,1,16) scanf("%s",s[i]+1); rep(i,1,16) rep(j,1,16) { num[i][j]=id(i,j); if(s[i][j]=='.') { A[++cnt]=(Point){i,j,0}; rep(k,1,16) if(s[i][k]!='.') A[cnt].s++; rep(k,1,16) if(s[k][j]!='.') A[cnt].s++; rep(k,1,16) rep(k2,1,16) if(s[k][k2]!='.'&&id(k,k2)==id(i,j)) A[cnt].s++; if(i==6||i==7||i==10) A[cnt].s+=200; } else { r[i][sc[s[i][j]]]=1;c[j][sc[s[i][j]]]=1; sq[num[i][j]][sc[s[i][j]]]=1; } } /*for(int i=1;i<cnt;i++) for(int j=i+1;j<=cnt;j++) if(A[j].x==6&&(A[i].x!=7)) swap(A[i],A[j]); else if(A[j].x==7) swap(A[i],A[j]); else if(A[j].x==10&&(A[i].x!=7)) swap(A[i],A[j]);*/ sort(A+1,A+cnt+1); //rep(i,1,cnt) printf("%d %d ",A[i].x,A[i].y); dfs(1); return 0; }
T3:
恩让我回忆回忆2333
-------------------
似乎记得要二分答案，那就先二分答案吧。
考虑如何判定x是否可行，将|E|/|V|>k变形成|E|-k*|V|>0，发现这是个裸的最大权闭合子图。因为选中一条边能贡献1的权，但同时要依赖于两个端点也必须选，而这两个端点的贡献每个都是-k。那么跑一边最大流，算一下是否为m即可。
等等，这道题要你输出分数啊！那么最后再计算一下最优解，在网络中找那些指向汇点且被割断的边，这些边的另一端就是选择的点。
注意特判m=0
#include<cstdio> #include<cstring> #include<cctype> #include<queue> #include<algorithm> #define rep(s,t) for(int i=s;i<=t;i++) #define ren for(int i=first[x];i!=-1;i=next[i]) using namespace std; inline int read() { char ch=getchar();int x=0,f=1; for(;!isdigit(ch);ch=getchar()) if(ch=='-') f=-1; for(;isdigit(ch);ch=getchar()) x=x*10+ch-'0'; return x*f; } const int maxn=2010; const int maxm=50100; const double eps=1e-5; int n,m,u[5010],v[5010],A[60],is[60],cnt,ans; struct Dinic { int n,m,s,t; int vis[maxn],cur[maxn],d[maxn],first[maxn],next[maxm]; struct Edge {int from,to;double flow;}edges[maxm]; void init(int n) { this->n=n;m=0; fill(first+1,first+n+1,-1); } void AddEdge(int u,int v,double cap) { edges[m]=(Edge){u,v,cap};next[m]=first[u];first[u]=m++; edges[m]=(Edge){v,u,0.0};next[m]=first[v];first[v]=m++; } int BFS() { fill(vis+1,vis+n+1,0); queue<int> q;q.push(s);d[s]=0;vis[s]=1; while(!q.empty()) { int x=q.front();q.pop();cur[x]=first[x]; ren { Edge& e=edges[i]; if(!vis[e.to]&&e.flow>eps) { vis[e.to]=1; d[e.to]=d[x]+1; q.push(e.to); } } } return vis[t]; } double DFS(int x,double a) { if(x==t||a<eps) return a; double flow=0.0,f; for(int& i=cur[x];i!=-1;i=next[i]) { Edge& e=edges[i]; if(d[e.to]==d[x]+1&&(f=DFS(e.to,min(a,e.flow)))>eps) { e.flow-=f;edges[i^1].flow+=f; flow+=f;a-=f;if(a<eps) break; } } return flow; } double MaxFlow(int s,int t) { double ans=0.0;this->s=s;this->t=t; while(BFS()) ans+=DFS(s,1e50); return ans; } void solve() { rep(0,m-1) if(edges[i].to==t&&edges[i].flow<eps) A[++cnt]=edges[i].from; } }sol; int check(double mid) { sol.init(n+m+2); int s=n+m+1,t=n+m+2; rep(1,m) sol.AddEdge(s,i,1.0),sol.AddEdge(i,u[i]+m,1e50),sol.AddEdge(i,v[i]+m,1e50); rep(1,n) sol.AddEdge(i+m,t,mid); return m-sol.MaxFlow(s,t)>eps; } int gcd(int a,int b) {return !b?a:gcd(b,a%b);} int main() { freopen("density.in","r",stdin); freopen("density.out","w",stdout); n=read(),m=read(); if(!m) {puts("0/1");return 0;} rep(1,m) u[i]=read(),v[i]=read(); double l=0.0,r=m; while(r-l>eps) { double mid=(l+r)*0.5; if(check(mid)) l=mid; else r=mid; } check(l);sol.solve(); rep(1,cnt) is[A[i]-m]=1; rep(1,m) if(is[u[i]]&&is[v[i]]) ans++; printf("%d/%d ",ans/gcd(ans,cnt),cnt/gcd(ans,cnt)); return 0; }
交上去WA了一个点，FHQ带大家测了半天，最后发现最小割不唯一会WA。于是暴力枚举分子分母的过了23333