今日SGU 5.27

SGU 122

题意：给你n个人，每个人有大于 N / 2(向上取整)的朋友，问你1这个人有一个书，每个人都想看，只能从朋友之间传递，然后最后回到了1这个人，问你

是否有解，然后有解输出路径

收获：哈密尔顿路

一:Dirac定理(充分条件)

　　设一个无向图中有N个顶点,若所有顶点的度数大于等于N/2,则哈密顿回路一定存在.(N/2指的是⌈N/2⌉,向上取整)

二:基本的必要条件

　　设图G=<V, E>是哈密顿图,则对于v的任意一个非空子集S,若以|S|表示S中元素的数目,G-S表示G中删除了S中的点以及这些点所关联的边后得到的子图,则W(G-S)<=|S|成立.其中W(G-S)是G-S中联通分支数.

三:竞赛图(哈密顿通路)

　　N(N>=2)阶竞赛图一点存在哈密顿通路.

还偷了一个哈密尔顿回路模板：https://blog.csdn.net/u010929036/article/details/46345059

而且这道题会卡输入的，我用getline超时了。。。

#include<bits/stdc++.h>

#define de(x) cout<<#x<<"="<<x<<endl;

#define dd(x) cout<<#x<<"="<<x<<" ";

#define rep(i,a,b) for(int i=a;i<(b);++i)

#define repd(i,a,b) for(int i=a;i>=(b);--i)

#define repp(i,a,b,t) for(int i=a;i<(b);i+=t)

#define ll long long

#define mt(a,b) memset(a,b,sizeof(a))

#define fi first

#define se second

#define inf 0x3f3f3f3f

#define INF 0x3f3f3f3f3f3f3f3f

#define pii pair<int,int>

#define pdd pair<double,double>

#define pdi pair<double,int>

#define mp(u,v) make_pair(u,v)

#define sz(a) (int)a.size()

#define ull unsigned long long

#define ll long long

#define pb push_back

#define PI acos(-1.0)

#define qc std::ios::sync_with_stdio(false)

#define db double

#define all(a) a.begin(),a.end()

const int mod = 1e9+;

const int maxn = 1e3+;

const double eps = 1e-;

using namespace std;

bool eq(const db &a, const db &b) { return fabs(a - b) < eps; }

bool ls(const db &a, const db &b) { return a + eps < b; }

bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); }

ll gcd(ll a,ll b) { return a==?b:gcd(b%a,a); };

ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }

ll kpow(ll a,ll b) {ll res=; if(b<) return ; for(;b;b>>=){if(b&)res=res*a;a=a*a;}return res;}

ll read(){

    ll x=,f=;char ch=getchar();

    while (ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while (ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

class Hamilton {

        int n, next[maxn];

        bool g[maxn][maxn], vis[maxn];

        int find(int u)

        {

                for (int v = ; v < n; ++v)

                        if (g[u][v] && !vis[v]) return v;

                return -;

        }

        void reverse(int v, int f)

        {

                if (v == -) return;

                reverse(next[v], v); next[v] = f;

        }

        public:

        void init(int n)

        {

                this->n = n;

                memset(g, false, sizeof(g));

        }

        void add_edge(int u, int v) { g[u][v] = true; }

        void find_path(int s = )

        {

                int t = s, sz = ;

                memset(next, -, sizeof(next));

                memset(vis, false, sizeof(vis)); vis[s] = true;

                while (sz < n) {

                        if (sz == ) {

                                for (int v; ~(v = find(s)); s = v)

                                        ++sz, vis[v] = true, next[v] = s;

                                for (int v; ~(v = find(t)); t = v)

                                        ++sz, vis[v] = true, next[t] = v;

                        } else {

                                for (int u, v = ; v < n; ++v) if (!vis[v]) {

                                        ++sz, vis[v] = true;

                                        for (u = s; !g[u][v]; u = next[u]);

                                        s = next[u]; t = next[u] = v; break;

                                }

                        }

                        if (g[t][s]) next[t] = s;

                        for (int u = next[s], v; next[t] == -; u = next[u])

                                if (g[u][t] && g[v=next[u]][s])

                                        reverse(v, s), next[u] = t, t = v;

                }

                for (int i = , u = ; i < n; ++i, u = next[u])

                        printf("%d ", u + );

                printf("%d\n", );

        }

} grp;

char s[],*p;

//s可以对应char *，而不能对应char* &

bool get_int(int &v,char* &p){

    v = ;

    while(*p && !isdigit(*p)) p++;

    if(!isdigit(*p)) return false;

    while(isdigit(*p)) v = v *  + *p++ - '';

    return true;

}

int main(){

    int n,v;

    scanf("%d",&n);getchar();

    grp.init(n);

    rep(i,,n){

        gets(s);p = s;

//        de(s)

        while(get_int(v,p)) grp.add_edge(i,--v);

    }

    grp.find_path();

    return ;

}

SGU  178

题意：题意自己了解下，打字不好描述

收获：打表模拟几组数据，暴力

#include<bits/stdc++.h>

#define de(x) cout<<#x<<"="<<x<<endl;

#define dd(x) cout<<#x<<"="<<x<<" ";

#define rep(i,a,b) for(int i=a;i<(b);++i)

#define repd(i,a,b) for(int i=a;i>=(b);--i)

#define repp(i,a,b,t) for(int i=a;i<(b);i+=t)

#define ll long long

#define mt(a,b) memset(a,b,sizeof(a))

#define fi first

#define se second

#define inf 0x3f3f3f3f

#define INF 0x3f3f3f3f3f3f3f3f

#define pii pair<int,int>

#define pdd pair<double,double>

#define pdi pair<double,int>

#define mp(u,v) make_pair(u,v)

#define sz(a) (int)a.size()

#define ull unsigned long long

#define ll long long

#define pb push_back

#define PI acos(-1.0)

#define qc std::ios::sync_with_stdio(false)

#define db double

#define all(a) a.begin(),a.end()

const int mod = 1e9+;

const int maxn = 2e2+;

const double eps = 1e-;

using namespace std;

bool eq(const db &a, const db &b) { return fabs(a - b) < eps; }

bool ls(const db &a, const db &b) { return a + eps < b; }

bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); }

ll gcd(ll a,ll b) { return a==?b:gcd(b%a,a); };

ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }

ll kpow(ll a,ll b) {ll res=; if(b<) return ; for(;b;b>>=){if(b&)res=res*a;a=a*a;}return res;}

ll read(){

    ll x=,f=;char ch=getchar();

    while (ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while (ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

ll n;

bool ok(int i){

    if(!i) return n==;

    ll sum = i,now = i + , cnt =  * i + ;

    if(sum >= n) return true;

    rep(j, , cnt - i){

        sum += now;

        now <<= ;

        if(sum >= n) return true;

    }

    return sum >= n;

}

int main(){

    scanf("%lld",&n);

    rep(i,,) if(ok(i)) return printf("%d\n",i),;

    return ;

}