dfs练习(二)

题目：Soldier and Cards

代码：

#include<bits/stdc++.h>
using namespace std;

const int maxn = 1e5+10;
queue<int>s, t;
map<queue<int>, queue<int> >mp;
int main(){
    int n, k1, k2;
    scanf("%d", &n);
    scanf("%d", &k1);
    for(int i = 1; i <= k1;i++){
        int v;
        scanf("%d", &v);
        s.push(v);
    }
    //if(mp[s] == t)cout<<"jkjk"<<endl;
    scanf("%d", &k2);

    for(int i = 1; i <= k2; i++){
        int v;
        scanf("%d", &v);
        t.push(v);
    }
    mp[s] = t;
    //cout<<s.size()<<" "<<t.size()<<endl;
    int flag = 0;
    int cn = 0;
    while(s.size() != n && s.size() != 0){
        int x, y;
        x = s.front();
        y = t.front();
        s.pop(), t.pop();
        if(x > y){
            s.push(y);
            s.push(x);
        }
        else {
            t.push(x);
            t.push(y);
        }
        cn++;
        if(mp.count(s) && mp[s] == t){//一定要使用mp.count,因为mp映射的队列默认位空队列，所以当s之前没有出现过，并且t位空的时候使结果出现问题
            flag = 1;
            break;
        }
        mp[s] = t;
    }
    if(flag)printf("-1\n");
    else {
            printf("%d ", cn);
            if(!s.size())printf("2\n");
            else printf("1\n");
    }


return 0;
}

Divide by three, multiply by two

代码：

#include<bits/stdc++.h>
using namespace std;

const int maxn = 200;
#define LL unsigned long long

LL a[maxn], mark[maxn], r[maxn];
map<LL, int>vis;
int n, cn;

int dfs(LL x, int step)
{
    if(step == n - 1)
    {
        cn = step;
        for(int i = 0; i < n; i++)
        {
            r[i] = mark[i];
        }
        return 1;
    }
    if(x % 3 == 0 && vis[x/3])
    {
        mark[step] = x/3;
        if(dfs(x/3, step+1))return 1;

    }

    if(vis[x*2])
    {
        mark[step] = x*2;
        if(dfs(x*2, step+1)) return 1;
    }
    return 0;
}
int main()
{
    scanf("%d", &n);
    vis.clear();
    for(int i = 1; i <= n; i++)
    {
        scanf("%llu", &a[i]);
        vis[a[i]]++;
    }
    for(int i = 1; i <= n; i++)
    {
        cn = 0;
        if(dfs(a[i], 0))
        {
            printf("%llu", a[i]);
            for(int i = 0; i < cn; i++)
            {
                printf(" %llu", r[i]);
            }
            printf("\n");
        }
    }

    return 0;
}

题目：Mahmoud and Ehab and the bipartiteness

代码：

#include<bits/stdc++.h>
using namespace std;
#define LL long long
const int maxn = 2e5+10;
int vis[maxn], n;
int tot = 1, head[maxn], Next[maxn], to[maxn];
void add(int u, int v){
     to[tot] = v;
     Next[tot] = head[u];
     head[u] = tot++;
}
void dfs(int x, int fa, int status){

     for(int i = head[x]; i; i = Next[i]){
        int v = to[i];
        if(v == fa)continue;
        if(status)vis[v] = 1;
        else vis[v] = 2;
        //cout<<v<<endl;
        dfs(v, x ,!status);
     }

}
int main(){
    scanf("%d", &n);
    tot = 1;
    for(int i = 1; i < n; i++){
        int u, v;
        scanf("%d %d", &u, &v);
        add(u, v);
        add(v, u);

    }
    int root;


    vis[1] = 1;
    dfs(1,-1,0);
    LL cn = 0;
    for(int i = 1; i <= n; i++){
        if(vis[i] == 1)cn++;
    }
    printf("%lld\n", (1LL)*cn*(n-cn)-n+1);

return 0;
}

题目： News Distribution

并查集
不知道用递推的方法为什么会超时，用递归的方法反而没有超时
代码：

#include<bits/stdc++.h>
using namespace std;

const int maxn = 5e5+10;
int f[maxn], sz[maxn], vis[maxn];
inline int read(){
   int s=0,w=1;
   char ch=getchar();
   while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
   while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();
   return s*w;
}
int Find(int x)//查询根
{
    if(f[x]==x)
        return x;
    else
        return f[x] = Find(f[x]);
}
void init(int n){
     for(int i = 1; i <= n; i++){
        sz[i] = 1;
        f[i] = i;
     }
}
int main(){
    int n, m;
    n = read(), m = read();
    init(n);
    for(int i = 1; i <= m; i++){
        int k, v, pre;
        k = read();
        for(int j = 1; j <= k; j++){
            v = read();
            if(j == 1){
                    pre = v;
                    continue;
            }
            int xx = Find(pre), yy = Find(v);
            if(xx == yy)continue;
            sz[xx] += sz[yy];
            f[yy] = xx;
        }
    }
    for(int i = 1; i <= n; i++){
        vis[i] = sz[Find(i)];
        printf("%d ", vis[i]);
    }
    printf("\n");

return 0;
}