【一本通】欧拉回路

欧拉回路

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+5,M=4e5+5;
int t,n,m,tot,in[N],ou[N],ans[M];
int cnt,fro[N],to[M],nxt[M];
bool vis[M];
void add(int x,int y) {
    to[++cnt]=y,nxt[cnt]=fro[x]; fro[x]=cnt;
    in[y]++,ou[x]++;
}

void DFS(int x) {
    for(int &i=fro[x];i;i=nxt[i]) {
        int h=i;
        if(t==1) {
            int k=(i+1)>>1;
            if(!vis[k]) {
                vis[k]=1; DFS(to[i]);
                if(h&1) ans[++tot]=k; else ans[++tot]=-k;
            }
        }
        else if(!vis[i]) {
            vis[i]=1; DFS(to[i]);
            ans[++tot]=h;
        }
    } 
}

int main() {
    t=read();
    n=read(),m=read();
    for(int i=1;i<=m;i++) {
        int u=read(),v=read();
        add(u,v); if(t==1) add(v,u);
    }
    if(t==1) for(int i=1;i<=n;i++) if(in[i]&1) return printf("NO"),0; 
    if(t==2) for(int i=1;i<=n;i++) if(in[i]^ou[i]) return printf("NO"),0;
    DFS(to[1]);
    if(tot<m) return printf("NO"),0;
    printf("YES
");
    for(int i=tot;i;i--) printf("%d ",ans[i]);
}

单词游戏

#include<bits/stdc++.h>
using namespace std;
int T,n,Father,in[30],out[30],fa[30];
bool flag,s,t;
char a[1005];

int num(char c) {return c-'a'+1;}
int getfa(int x) {return fa[x]==x?x:fa[x]=getfa(fa[x]);}
void ready() {
    flag=s=t=false; Father=0;
    for(int i=1;i<=26;i++) fa[i]=i;
    memset(in,0,sizeof(in));
    memset(out,0,sizeof(out));
}

int main() {
    scanf("%d",&T);
    while(T--) {
        ready();
        scanf("%d",&n);
        for(int i=1;i<=n;i++) {
            scanf("%s",a+1);
            int len=strlen(a+1);
            out[num(a[1])]++,in[num(a[len])]++;
            fa[getfa(num(a[1]))]=getfa(num(a[len]));
        }
        for(int i=1;i<=26;i++) {
            if(in[i]==out[i]) continue;
            else if(in[i]+1==out[i]&&!s) s=true;
            else if(in[i]==out[i]+1&&!t) t=true;
            else {flag=true; break;}
        }
        for(int i=1;i<=26;i++)
         if(in[i]||out[i]) {
             if(!Father) Father=getfa(i);
             else if(Father^getfa(i)) {flag=true; break;}
         }
        if(flag) printf("The door cannot be opened.
");
        else printf("Ordering is possible.
");
    }
}

Ant Trip

题解

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+5;
int n,m,ans,d[N],fa[N],s[N];
bool vis[N];
vector<int> v;

int getfa(int x) {return fa[x]==x?x:fa[x]=getfa(fa[x]);}
void ready() {
    ans=0;
    v.clear();
    memset(d,0,sizeof(d));
    memset(vis,0,sizeof(vis));
    memset(s,0,sizeof(s));
    for(int i=1;i<=n;i++) fa[i]=i;
}

int main() {
    while(~scanf("%d%d",&n,&m)) {
        ready();
        for(int i=1;i<=m;i++) {
            int a=read(),b=read();
            d[a]++,d[b]++;
            fa[getfa(a)]=getfa(b);
        }
        for(int i=1;i<=n;i++) {
            int x=getfa(i);
            if(!vis[x]) {
                v.push_back(x);
                vis[x]=1;
            }
            if(d[i]&1) s[x]++;
        }
        for(int i=0;i<v.size();i++) {
            int x=v[i];
            if(d[x]==0) continue;
            if(s[x]==0) ans++;
            else ans+=s[x]/2;
        }
        printf("%d
",ans);
    }
}