A:签到。
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #include<algorithm> using namespace std; int gcd(int n,int m){return m==0?n:gcd(m,n%m);} int read() { int x=0,f=1;char c=getchar(); while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();} while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),c=getchar(); return x*f; } int n,a,b; int main() { /*freopen("a.in","r",stdin); freopen("a.out","w",stdout);*/ n=read(),a=read(),b=read(); cout<<min(a,b)<<‘ ‘<<max(0,a+b-n); return 0; }
B:签到。
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #include<algorithm> using namespace std; int gcd(int n,int m){return m==0?n:gcd(m,n%m);} int read() { int x=0,f=1;char c=getchar(); while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();} while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),c=getchar(); return x*f; } #define N 110 int n; char a[N],b[N],c[N]; int main() { /*freopen("a.in","r",stdin); freopen("a.out","w",stdout);*/ n=read(); scanf("%s",a+1);scanf("%s",b+1);scanf("%s",c+1); int ans=0; for (int i=1;i<=n;i++) { if (a[i]==b[i]&&b[i]==c[i]) ; else if (a[i]==b[i]||a[i]==c[i]||b[i]==c[i]) ans++; else ans+=2; } cout<<ans; return 0; }
C:考虑一种菜自己吃和对方吃的收益差,于是显然按ai+bi排序从大到小选即可。
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #include<algorithm> using namespace std; int gcd(int n,int m){return m==0?n:gcd(m,n%m);} int read() { int x=0,f=1;char c=getchar(); while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();} while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),c=getchar(); return x*f; } #define N 300010 #define ll long long int n; struct data { int x,y; bool operator <(const data&a) const { return x+y<a.x+a.y; } }a[N]; ll ans; int main() { /*freopen("a.in","r",stdin); freopen("a.out","w",stdout);*/ n=read(); for (int i=1;i<=n;i++) a[i].x=read(),a[i].y=read(); sort(a+1,a+n+1);reverse(a+1,a+n+1); for (int i=1;i<=n;i++) if (i&1) ans+=a[i].x; else ans-=a[i].y; cout<<ans; return 0; }
D:显然图仍是一个DAG,其中度数为0的点是原树的根。由于图中没有重边,瞎考虑一下容易发现,对于每一个点,由根到它的最长路上该点的前驱即为其在原树中的父亲。拓扑排序一下即可。开始写了个不知道啥玩意于是比E晚了10min。
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #include<algorithm> using namespace std; int gcd(int n,int m){return m==0?n:gcd(m,n%m);} int read() { int x=0,f=1;char c=getchar(); while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();} while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),c=getchar(); return x*f; } #define N 300010 #define ll long long int n,m,p[N],fa[N],deep[N],degree[N],q[N],t,root; bool flag[N]; struct data{int to,nxt; }edge[N]; void addedge(int x,int y){t++;edge[t].to=y,edge[t].nxt=p[x],p[x]=t;} void topsort() { int head=0,tail=0; for (int i=1;i<=n;i++) if (!degree[i]) q[++tail]=i; while (tail<n) { int x=q[++head]; for (int i=p[x];i;i=edge[i].nxt) { degree[edge[i].to]--; if (deep[x]+1>deep[edge[i].to]) { deep[edge[i].to]=deep[x]+1; fa[edge[i].to]=x; } if (!degree[edge[i].to]) q[++tail]=edge[i].to; } } } int main() { n=read(),m=read(); for (int i=1;i<n+m;i++) { int x=read(),y=read(); addedge(x,y);degree[y]++; } topsort(); for (int i=1;i<=n;i++) printf("%d\n",fa[i]); return 0; }
E:显然从大到小考虑每条边是否需要删即可,但删边的过程中难以维护连通块信息。注意到只有MST的边会对最后所得图的连通性产生影响。于是求出MST,然后可以从大到小删边用LCT维护,但这样显然比较毒瘤。事实上还可以建出kruskal重构树,从大到小考虑每条边,如果其在重构树的父亲不需要被删掉,显然其也不需要被删掉;否则此时其所在连通块即为其在重构树上的子树,判断一下是否需要删掉即可。这样就得到了最后的连通信息,最后再考虑每条边是否保留即可。
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #include<algorithm> using namespace std; int gcd(int n,int m){return m==0?n:gcd(m,n%m);} int read() { int x=0,f=1;char c=getchar(); while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();} while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),c=getchar(); return x*f; } #define N 400010 #define ll long long int n,m,a[N],fa[N],id[N],cnt,ans; bool isdel[N],flag[N]; ll value[N<<2]; struct data { int x,y,z; bool operator <(const data&a) const { return z<a.z; } }e[N]; int find(int x){return fa[x]==x?x:fa[x]=find(fa[x]);} namespace kruskal_tree { int p[N<<2],t,degree[N<<2],fa[N<<2]; ll size[N<<2]; struct data{int to,nxt;}edge[N<<2]; void addedge(int x,int y){t++;degree[y]++;fa[y]=x;edge[t].to=y,edge[t].nxt=p[x],p[x]=t;} void link(int x,int y,int i) { id[i]=++cnt;addedge(cnt,x),addedge(cnt,y); } void dfs(int k) { size[k]=a[k]; for (int i=p[k];i;i=edge[i].nxt) { dfs(edge[i].to); size[k]+=size[edge[i].to]; } } void build() { for (int i=1;i<=cnt;i++) if (!degree[i]) dfs(i); } void del() { isdel[0]=1; for (int i=m;i>=1;i--) if (flag[i]&&isdel[fa[id[i]]]&&e[i].z>size[id[i]]) isdel[id[i]]=1; } } void dfs(int k) { for (int i=kruskal_tree::p[k];i;i=kruskal_tree::edge[i].nxt) { dfs(kruskal_tree::edge[i].to); if (!isdel[k]) fa[find(kruskal_tree::edge[i].to)]=k; } } int main() { n=read(),m=read(); for (int i=1;i<=n;i++) a[i]=read(); for (int i=1;i<=m;i++) e[i].x=read(),e[i].y=read(),e[i].z=read(); sort(e+1,e+m+1); for (int i=1;i<=4*n;i++) fa[i]=i;cnt=n; for (int i=1;i<=m;i++) { int p=find(e[i].x),q=find(e[i].y); if (p!=q) kruskal_tree::link(p,q,i),fa[p]=fa[q]=id[i],flag[i]=1; } kruskal_tree::build(); kruskal_tree::del(); for (int i=1;i<=cnt;i++) fa[i]=i; for (int i=1;i<=cnt;i++) if (!kruskal_tree::degree[i]) dfs(i); for (int i=1;i<=m;i++) if (find(e[i].x)==find(e[i].y)&&kruskal_tree::size[find(e[i].x)]>=e[i].z) ans++; cout<<m-ans; return 0; }
result:rank 132 rating +84
原文地址:https://www.cnblogs.com/Gloid/p/10328307.html
时间: 2024-10-08 03:33:44