2017ACM萧山训练第3场(World Final 2013)
A.Self-Assembly
如果一个正方形有两条边a,b
则a->op(b) b->op(a)，判图中是否有环，有环则说明我们能把一些正方形绕成环然后翻折旋转变得无限大
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#include<map> #include<cmath> #include<ctime> #include<queue> #include<cstdio> #include<vector> #include<bitset> #include<cstring> #include<iostream> #include<algorithm> #define pa pair<int,int> #define ll long long #define inf 1000000000 using namespace std; 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; } int n; int a[5],mark[60],mp[60][60]; int getint(char a,char b) { if(a=='0')return -1; int t=a-'A'; if(b=='-')t+=26; return t; } int op(int x) { return x<26?x+26:x-26; } void dfs(int x) { for(int i=0;i<52;i++) if(!mark[i]&&mp[x][i]) { mark[i]=1; dfs(i); } } int main() { n=read(); char ch[10]; for(int i=1;i<=n;i++) { scanf("%s",ch+1); for(int k=1;k<=4;k++) a[k]=getint(ch[k*2-1],ch[k*2]); for(int j=1;j<=4;j++) for(int k=j+1;k<=4;k++) if(a[j]!=-1&&a[k]!=-1) { int u=a[j],v=a[k]; mp[u][op(v)]=mp[v][op(u)]=1; } } for(int i=0;i<52;i++) { memset(mark,0,sizeof(mark)); dfs(i); if(mark[i]){puts("unbounded");return 0;} } puts("bounded"); return 0; } |
C.Surely You Congest
只有最短路相同的会互相影响
按最短路分组后跑 c 次最大流
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#include<map> #include<cmath> #include<ctime> #include<queue> #include<cstdio> #include<vector> #include<bitset> #include<cstring> #include<iostream> #include<algorithm> #define pa pair<int,int> #define ll long long #define inf 1000000000 using namespace std; 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; } int n,m,c,cnt,ans; int last[25005],dis[25005],q[25005],h[25005],id[25005],cur[25005]; int u[50005],v[50005],w[50005]; bool vis[25005]; struct edge{ int to,next,v; }e[200005]; priority_queue<pa,vector<pa>,greater<pa> >pq; void insert(int u,int v,int w) { e[++cnt]=(edge){v,last[u],w};last[u]=cnt; } void dijkstra() { pq.push(make_pair(0,1)); dis[1]=0;for(int i=2;i<=n;i++)dis[i]=inf; while(!pq.empty()) { int x=pq.top().second;pq.pop(); if(vis[x])continue;vis[x]=1; for(int i=last[x];i;i=e[i].next) if(!vis[e[i].to]&&dis[x]+e[i].v<dis[e[i].to]) { dis[e[i].to]=dis[x]+e[i].v; pq.push(make_pair(dis[e[i].to],e[i].to)); } } } bool bfs() { int head=0,tail=1; for(int i=0;i<=n;i++)h[i]=-1; q[0]=0;h[0]=0; while(head!=tail) { int x=q[head];head++; for(int i=last[x];i;i=e[i].next) if(h[e[i].to]==-1&&e[i].v) { h[e[i].to]=h[x]+1; q[tail++]=e[i].to; } } return h[1]!=-1; } int dfs(int x,int f) { if(x==1)return f; int w,used=0; for(int i=cur[x];i;i=e[i].next) if(h[x]+1==h[e[i].to]&&e[i].v) { w=dfs(e[i].to,min(e[i].v,f-used)); e[i].v-=w;e[i^1].v+=w; if(e[i].v)cur[x]=i; used+=w;if(used==f)return f; } if(!used)h[x]=-1; return used; } void solve(int a,int b) { cnt=1;for(int i=0;i<=n;i++)last[i]=0; int tmp = 0; for(int i=a;i<=b;i++) if(dis[id[i]]<inf) { insert(0,id[i],1); insert(id[i],0,0); tmp++; } if(tmp<=1) { ans+=tmp; return; } for(int i=1;i<=m;i++) { int a=u[i],b=v[i],c=w[i]; insert(b,a,1); insert(a,b,0); } while(bfs()) { for(int i=0;i<=n;i++)cur[i]=last[i]; ans+=dfs(0,inf); } } bool cmp(int a,int b) { return dis[a]<dis[b]; } int main() { n=read();m=read();c=read(); for(int i=1;i<=m;i++) { u[i]=read(),v[i]=read(),w[i]=read(); insert(u[i],v[i],w[i]); insert(v[i],u[i],w[i]); } for(int i=1;i<=c;i++)id[i]=read(); dijkstra(); int t=0; for(int i=1;i<=m;i++) { if(dis[u[i]]>dis[v[i]])swap(u[i],v[i]); if(dis[u[i]]+w[i]==dis[v[i]]) { t++; swap(u[t],u[i]); swap(v[t],v[i]); swap(w[t],w[i]); } } m=t; sort(id+1,id+c+1,cmp); for(int i=1,j;i<=c;i=j) { for(j=i;dis[id[j]]==dis[id[i]]&&j<=c;j++); solve(i,j-1); } printf("%d\n",ans); return 0; } |
D.Factors
爆搜前16个素数
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#include<map> #include<cmath> #include<ctime> #include<queue> #include<cstdio> #include<vector> #include<bitset> #include<cstring> #include<iostream> #include<algorithm> #define pa pair<int,int> #define ll unsigned long long #define inf (1ULL<<63) using namespace std; ll n; ll pri[]={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53}; ll c[65][65]; map<ll,ll>ans; void pre() { c[0][0]=1; for(int i=1;i<=62;i++) { c[i][0]=1; for(int j=1;j<=i;j++) if(c[i-1][j-1]<inf&&c[i-1][j]<inf) c[i][j]=c[i-1][j]+c[i-1][j-1]; else c[i][j]=inf; } } void dfs(ll n,ll now,ll tot,int k) { if(tot>0) if(ans[n]==0||now<ans[n]) ans[n]=now; ll t=now,cnt=0; while(1) { if(pri[k]>inf/t) break; t*=pri[k];cnt++; if(c[tot+cnt][cnt]<=inf/n) dfs(n*c[tot+cnt][cnt],t,tot+cnt,k+1); } } int main() { pre(); dfs(1,1,0,0); while(scanf("%llu",&n)!=EOF) { printf("%llu %llu \n",n,ans[n]); } return 0; } |
F.Low Power
二分答案贪心检验
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#include<cmath> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define ll long long using namespace std; 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; } int n,K,m; int a[1000005]; bool mark[1000005]; bool check(int mid) { memset(mark,0,sizeof(mark)); int now=0,t=0,p=0; for(int i=1;i<m;i++) if(!mark[i]&&now<n&&a[i+1]-a[i]<=mid) { mark[i]=mark[i+1]=1; now++; } if(now!=n)return 0; for(int i=m;i;i--) if(!mark[i])t++; else { p++; if(t<p*(K-1))return 0; } return 1; } int main() { n=read();K=read();m=2*n*K; for(int i=1;i<=m;i++)a[i]=read(); sort(a+1,a+m+1); int l=0,r=1000000000; while(l<=r) { int mid=(l+r)>>1; if(check(mid))r=mid-1; else l=mid+1; } printf("%d\n",l); return 0; } |
H:Матрëшка
预处理\(f(i,j)\)表示将 i 到 j 全部套在一起的最小代价，枚举断点
合并两组套娃，不用拆开的情况是某一个套娃比另一组的最小套娃还小
维护前缀和，前缀后缀的最小值，能够\(O(1)\)完成转移
\(g(i)\)表示把前 i 个套娃套成若干完整套娃的最小代价，转移显然
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#include <cmath> #include <cstdio> #include <cstdlib> #include <iostream> #include <algorithm> using namespace std; int v[505],sum[505][505],LG[505],f[505][11],n,a[505],i,j,M1,M2,dp[505][505],k,DP[505],sum2,MAX; int MIN(int l,int r) { int t=LG[r-l+1]; return min(f[l][t],f[r-(1<<t)+1][t]); } int main() { scanf("%d",&n); for (i=1; i<=n; i++) scanf("%d",&a[i]); for (i=1; i<=n; i++) sum[i][a[i]]=1; for (i=1; i<=n; i++) for (j=1; j<=500; j++) sum[i][j]+=sum[i-1][j]; for (i=1; i<=n; i++) for (j=1; j<=500; j++) sum[i][j]+=sum[i][j-1]; for (i=1; i<=n; i++) f[i][0]=a[i]; for (i=1; i<=n; i++) LG[i]=int(log(i)/log(2)+0.000000001); for (i=1; i<=LG[n]; i++) for (j=1; j<=n-(1<<i)+1; j++) f[j][i]=min(f[j][i-1],f[j+(1<<i-1)][i-1]); for (i=1; i<n; i++) for (j=1; j<=n-i; j++) { dp[j][i+j]=453266144; for (k=j; k<i+j; k++) { M1=MIN(j,k); M2=MIN(k+1,i+j); dp[j][i+j]=min(dp[j][i+j],dp[j][k]+dp[k+1][i+j]+i+1-sum[i+j][M1]+sum[k][M1]-sum[k][M2]+sum[j-1][M2]); } } for (i=1; i<=n; i++) { DP[i]=453266144; MAX=0; sum2=0; for (j=i; j>=1; j--) { if (v[a[j]]==i) break; v[a[j]]=i; MAX=max(MAX,a[j]); sum2+=a[j]; if (MAX*(MAX+1)/2==sum2) DP[i]=min(DP[i],DP[j-1]+dp[j][i]); } } if (DP[n]>=453266144) cout<<"Impossible";else cout<<DP[n]; return 0; } |
I.Pirate Chest
用单调栈把暴力优化成\(O(n^3)\)
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#include<map> #include<set> #include<cmath> #include<ctime> #include<queue> #include<cstdio> #include<vector> #include<bitset> #include<cstring> #include<iostream> #include<algorithm> #define inf 1000000000 #define ll long long using namespace std; 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; } ll ans; ll a,b,m,n; int h[505][505],val[505]; int q[505]; int l[505],r[505],mn[505][505]; void solve(int x,int y) { if(y>a&&y>b)return; int lim=max(a,b); if(y>a)lim=a; else if(y>b)lim=b; for(int k=1;k<=m;k++)val[k]=mn[y][k]=min(mn[y-1][k],h[x][k]); int top=0;val[0]=-1; for(int i=1;i<=m;i++) { while(val[i]<=val[q[top]])top--; l[i]=q[top];q[++top]=i; } top=0;q[++top]=m+1;val[m+1]=-1; for(int i=m;i;i--) { while(val[i]<=val[q[top]])top--; r[i]=q[top];q[++top]=i; } for(int i=1;i<=m;i++) { ll len=min(lim,r[i]-l[i]-1); ll h=(n*m*val[i]-1)/(n*m-y*len); ans=max(ans,y*h*len); } } int main() { memset(mn,127,sizeof(mn)); a=read();b=read();n=read();m=read(); for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) h[i][j]=read(); for(int i=n;i;i--) for(int j=n-i+1;j;j--) solve(i,j); printf("%lld\n",ans); return 0; } |
J.Pollution Solution
辛普森积分
但是可能被构造卡掉，可以把积分的范围调整一下
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#include<map> #include<cmath> #include<ctime> #include<queue> #include<cstdio> #include<vector> #include<bitset> #include<cstring> #include<iostream> #include<algorithm> #define eps 1e-8 #define pa pair<int,int> #define ll long long #define inf 1000000000 using namespace std; struct P{ double x, y; P(){} P(double _x, double _y):x(_x), y(_y){} friend P operator-(P a, P b){ return P(a.x - b.x, a.y - b.y); } friend double operator*(P a, P b){ return a.x * b.y - a.y * b.x; } }p[105]; struct L{ P a, b; }l[105]; vector<double> ve; int n; double R; void add(double X, L l) { if(l.a.x < X - eps && l.b.x < X - eps)return; if(l.a.x > X + eps && l.b.x > X + eps)return; if(fabs(X - l.a.x) < eps && fabs(X - l.b.x) < eps)return; double t; if(fabs(l.b.x - X) < eps)t = l.b.y; else t = (X - l.a.x) / (l.b.x - X); ve.push_back(l.a.y + (l.b.y - l.a.y) * t / (t + 1)); } double F(double X) { if(abs(X) > R)return 0; ve.clear(); for(int i = 1; i <= n; i++) add(X, l[i]); double U = sqrt(R * R - X * X), ANS = 0; sort(ve.begin(), ve.end()); bool flag = 0; for(int i = 0; i < (int)ve.size() - 1; i++) { flag ^= 1; if(ve[i] > U)break; if(flag) ANS += min(ve[i + 1], U) - ve[i]; } return ANS; } double cal(double fl, double fm, double fr, double L) { return (fl + fm * 4 + fr) * L / 6; } double simpson(double l, double r, double fl, double fm, double fr, double S, int dep) { double mid = (l + r) / 2; double m1 = (l + mid) / 2, m2 = (r + mid) / 2; double f1 = F(m1), f2 = F(m2); double g1 = cal(fl, f1, fm, mid - l), g2 = cal(fm, f2, fr, r - mid); if(fabs(g1 + g2 - S) < 1e-12 && dep >= 10)return g1 + g2; return simpson(l, mid, fl, f1, fm, g1, dep + 1) + simpson(mid, r, fm, f2, fr, g2, dep + 1); } double solve(double l, double r) { if(l > r)return 0; double mid = (l + r) / 2; return simpson(l, r, 0, F(mid), 0, (F(l) + F(r) + F(mid) * 4) * (r - l) / 6, 0); } int main() { scanf("%d%lf", &n, &R); for(int i = 1; i <= n; i++) scanf("%lf%lf", &p[i].x, &p[i].y); p[n + 1] = p[1]; for(int i = 1; i <= n; i++) l[i].a = p[i], l[i].b = p[i + 1]; double ANS = solve(-R, -2) + solve(-2, 1) + solve(1, R); printf("%.10lf\n", ANS); return 0; } |
计数器 9850951
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又是一年WF时