/* 有谁知道这道题结论是怎么来的？ 晚上问问学数学的孩子23333 */ #include<iostream> #include<cstdio> #include<cstdlib> #include<cmath> #include<cstring> using namespace std; typedef long long LL; LL n,m,d; LL gcd(LL x,LL y) { return x%y?gcd(y,x%y):y; } int main() { freopen("line.in","r",stdin); freopen("line.out","w",stdout); cin>>n>>m; d=gcd(n,m); n/=d; m/=d; if((n+m)&1)cout<<"1/2"<<endl; else cout<<(n*m+1)/2<<"/"<<n*m<<endl; return 0; }
/* 数据水... */ #include<iostream> #include<cstdio> #include<cstring> #define N 100007 using namespace std; int n,m,opt,x,y,z,ans,cnt; int col[N]; inline int read() { int x=0,f=1;char c=getchar(); while(c>'9'||c<'0'){if(c=='-')f=-1;c=getchar();} while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();} return x*f; } inline void solve(int l,int r,int flag) { for(int i=l;i<=r;i++) col[i]=flag; } inline void find(int x) { int L=1,R=n; if(col[x]==1) printf("0\n"); else { for(int i=x-1;i>=1;i--) if(col[i]){L=i+1;break;} for(int i=x+1;i<=n;i++) if(col[i]){R=i-1;break;} if(L==1 || R==n){printf("INF\n");} else printf("%d\n",R-L+1); } } int main() { freopen("explore.in","r",stdin); freopen("explore.out","w",stdout); n=read();m=read(); for(int i=1;i<=m;i++) { opt=read(); if(opt==1) { x=read();y=read(); solve(x,y,1); } else if(opt==2) { x=read();y=read(); solve(x,y,0); } else x=read(),find(x); } return 0; }
/*
二分+线段树 By LXT 林欣彤 */ #include<complex> #include<cstdio> using namespace std; const int N=1e5+7; int n,m; int Fog[N<<2],lazy[N<<2]; int qread() { int x=0; char ch=getchar(); while(ch<'0' || ch>'9')ch=getchar(); while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();} return x; } void PushUp(int rt) { Fog[rt]=(Fog[rt<<1]==1) || (Fog[rt<<1|1]==1); } void PushDown(int rt) { if(lazy[rt]) { Fog[rt<<1]=Fog[rt<<1|1]=lazy[rt]; lazy[rt<<1]=lazy[rt<<1|1]=lazy[rt]; lazy[rt]=0; } } void Modify(int l,int r,int rt,int nowl,int nowr,int v) { if(nowl<=l && r<=nowr) { Fog[rt]=lazy[rt]=v; return; } int mid=l+r>>1; PushDown(rt); if(nowl<=mid)Modify(l,mid,rt<<1,nowl,nowr,v); if(mid<nowr)Modify(mid+1,r,rt<<1|1,nowl,nowr,v); PushUp(rt); } int Query(int l,int r,int rt,int nowl,int nowr) { if(nowl<=l && r<=nowr)return Fog[rt]; int mid=l+r>>1; PushDown(rt); if(nowl<=mid)if(Query(l,mid,rt<<1,nowl,nowr)==1)return 1; if(mid<nowr)if(Query(mid+1,r,rt<<1|1,nowl,nowr)==1)return 1; return 2; } int main() { freopen("explore.in","r",stdin); freopen("explore.out","w",stdout); scanf("%d%d",&n,&m); int p,l,r; while(m--) { p=qread();l=qread(); if(p==3) { if(Query(1,n,1,l,l)==1){puts("0");continue;} int L=1,R=l,res1=l,res2=l; while(L<=R) { int mid=L+R>>1; if(Query(1,n,1,mid,l)==2) res1=mid,R=mid-1; else L=mid+1; } L=l;R=n; while(L<=R) { int mid=L+R>>1; if(Query(1,n,1,l,mid)==2) res2=mid,L=mid+1; else R=mid-1; } if(res1==1 || res2==n)puts("INF"); else printf("%d\n",res2-res1+1); } else { r=qread(); Modify(1,n,1,l,r,p); } } fclose(stdin);fclose(stdout); return 0; }
#include<complex> #include<cstdio> #include<iostream> using namespace std; const int N=1e5+7; struct node{ int u,v,w,nxt; }e[N<<1]; int n,m,Enum,tim,tot; long long ans=1ll<<60; int front[N],in[N],b[N],c[N],fat[N],Log[N],Smin[N][18]; bool a[N]; int qread() { int x=0; char ch=getchar(); while(ch<'0' || ch>'9')ch=getchar(); while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();} return x; } void Insert(int u,int v,int w) { e[++Enum].u=u; e[Enum].v=v; e[Enum].w=w; e[Enum].nxt=front[u]; front[u]=Enum; } int find(int x) { if(fat[x]!=x)fat[x]=find(fat[x]); return fat[x]; } void dfs1(int num) { if(num==n-1) { long long res=0; for(int i=1;i<n;i++) res+=b[i]*e[i<<1].w; for(int i=1;i<=n;i++) fat[i]=i; for(int i=1;i<n;i++) if(!b[i]) { int r1=find(e[i<<1].u),r2=find(e[i<<1].v); fat[r2]=r1; } for(int i=1;i<=m;i++) for(int j=i+1;j<=m;j++) if(find(c[i])==find(c[j]))return; ans=min(ans,res); return; } b[num+1]=0;dfs1(num+1); b[num+1]=1;dfs1(num+1); } void Solve1() { dfs1(0); cout<<ans<<endl; } void dfs(int x,int from,int w) { Smin[++tim][0]=w; if(a[x])b[++tot]=tim; for(int i=front[x];i;i=e[i].nxt) { int v=e[i].v; if(v==from)continue; dfs(v,x,e[i].w); } } int Query(int l,int r) { int k=Log[r-l+1]; return min(Smin[l][k],Smin[r-(1<<k)+1][k]); } void ST() { for(int i=2;i<=tim;i++) Log[i]=Log[i>>1]+1; for(int j=1;j<=Log[tim];j++) for(int i=tim-(1<<j-1);i;i--) Smin[i][j]=min(Smin[i][j-1],Smin[i+(1<<j-1)][j-1]); } void Solve2() { for(int i=1;i<=n;i++) if(in[i]==1) { dfs(i,0,0x3f3f3f3f); break; } ST(); long long res=0; for(int i=2;i<=tot;i++) res+=Query(b[i-1]+1,b[i]); cout<<res<<endl; } int main() { // freopen("apple.in","r",stdin); // freopen("apple.out","w",stdout); scanf("%d%d",&n,&m); int u,v,w; for(int i=1;i<n;i++) { u=qread()+1;v=qread()+1;w=qread(); Insert(u,v,w);Insert(v,u,w); in[u]++;in[v]++; } for(int i=1;i<=m;i++) c[i]=qread()+1,a[c[i]]=1; int k=0; for(int i=1;i<=n;i++) if(in[i]==2)k++; if(k==n-2)Solve2(); else Solve1(); fclose(stdin);fclose(stdout); return 0; }
/* 形 DP。首先任取一个有苹果的结点为根，构建整棵树，接下来自底向上进行动态 规划。设 f[u]表示以 u 为根的子树中苹果互不连通，且子树中的苹果不与树外的苹果连通的 最小代价。g[u]表示以 u 为根的子树中苹果互不连通所需的最小代价 那么当 u 上有苹果时， g[u] = sum (f[v]) （v 为 u 的所有子节点） ； f[u] = g[u] + pre[u] (pre[u]为 u 与其父节点连边的权值） 。 当 u 上无苹果时， 设 s[u] = sum (f[v]) （v 为 u 的所有子节点） 则 g[u] = min (s[u] - f[v] + g[v]) f[u] = min(s[u], g[u] + pre[u]) */ #include<cstdio> #include<algorithm> #include<cstring> #include<iostream> #include<queue> #define ll long long #define M 200010 using namespace std; const ll inf = 10000000000000000ll; ll f[M], g[M], p[M]; ll nxt[M], head[M], ver[M], to[M], cnt, n, k, sz[M]; bool is[M]; int sta[M]; int read() { int nm = 0, f = 1; char c = getchar(); for(; !isdigit(c); c = getchar()) if(c == '-') f = -1; for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0'; return nm * f; } void push(int vi, int vj, int v) { cnt++, nxt[cnt] = head[vi], head[vi] = cnt, to[cnt] = vj, ver[cnt] = v; } void dfs1(int now, int fa) { if(is[now]) sz[now] = 1; for(int i = head[now]; i; i = nxt[i]) { int vj = to[i]; if(vj == fa) continue; dfs1(vj, now); sz[now] += sz[vj]; } } void dfs2(int now, int fa) { if(sz[now]) f[now] = g[now] = inf; ll tot = 0; for(int i = head[now]; i; i = nxt[i]) { int vj = to[i]; if(vj == fa) continue; if(sz[vj]) { dfs2(vj, now); tot += g[vj]; } } ll tot2 = 0, maxx = 0; for(int i = head[now]; i; i = nxt[i]) { int vj = to[i]; if(vj == fa || !sz[vj]) continue; ll op = min(ver[i], f[vj] - g[vj]); tot2 += op; maxx = max(maxx, op); } if(is[now]) f[now] = inf, g[now] = tot + tot2; else f[now] = tot + tot2, g[now] = tot + tot2 - maxx; } int main() { freopen("apple.in", "r", stdin); freopen("apple.out", "w", stdout); n = read(), k = read(); for(int i = 1; i < n; i++) { int vi = read(), vj = read(), v = read(); push(vi, vj, v); push(vj, vi, v); } for(int i = 1; i <= k; i++) is[read()] = true; dfs1(0, 0); dfs2(0, 0); cout << min(f[0], g[0]) << "\n"; return 0; }