今日SGU 5.20

SGU 404

题意：。。

收获：取模

#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 = 1e2+;

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=;a%=mod; if(b<) return ; for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}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;

}

//inv[1]=1;

//for(int i=2;i<=n;i++) inv[i]=(mod-mod/i)*inv[mod%i]%mod;

int main(){

    string s[maxn];

    int n,m;

    cin>>n>>m;

    rep(i,,m+) cin>>s[i];

    cout<<s[n%m==?m:n%m];

    return ;

}

SGU 404

题意：给你n个人，每个人都有si和bi，两个人会打架（si>=sj&&bi<=bj || si<=sj&&bi>=bj）,然后让你邀请最多的人数

里面的所有人都不能打架

收获：反过来想，就是两个人能同时出现的就是其中一个人的s和b都严格大于另一个人的s和b，那我们排完序之后就是

一个lis的问题了

下面是一个我wa了很久的代码，

反例：

5
1 1
2 9
3 4
5 5
1 1

#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 = 1e5+;

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=;a%=mod; if(b<) return ; for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}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;

}

//inv[1]=1;

//for(int i=2;i<=n;i++) inv[i]=(mod-mod/i)*inv[mod%i]%mod;

int n;

struct P{

    int s,b,id;

    bool operator <(const P& c) const{

        if(s==c.s) return b>c.b;

        return s<c.s;

    }

}p[maxn];

int g[maxn],pre[maxn];

void init(){

    mt(g,);mt(pre,-);g[] = -;

}

void output(int x){

    if(pre[x]==-){

        printf("%d",p[x].id);

        return;

    }

    output(pre[x]);

    printf(" %d",p[x].id);

}

int main(){

    init();

    scanf("%d",&n);

    rep(i,,n+) scanf("%d%d",&p[i].s,&p[i].b),p[i].id = i;

    sort(p+,p+n+);

    int ans = ;

    rep(i,,n+){

        int l = ,r = ans,pos = -;

        while(l <= r){ //在前面找出找到第一个大于等于当前的b的点

            int mid = (l + r) >> ;

            if(p[i].b > p[g[mid]].b) l = mid + ;

            else pos = mid,r = mid - ;

        }

        if(pos == -) ans++,pos = ans;

        pre[i] = g[pos - ];

        if(p[i].s <= p[g[pos]].s || g[pos] == ) g[pos] = i;

    }

    printf("%d\n",ans);

    output(g[ans]);

    return ;

}

因为s已经是递增的了，那么后面的s肯定是大于等于前面的了，那我们只要让每个s对应的b最小，这样后面的才更容易拼起来

AC代码：

#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 = 1e5+;

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=;a%=mod; if(b<) return ; for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}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;

}

//inv[1]=1;

//for(int i=2;i<=n;i++) inv[i]=(mod-mod/i)*inv[mod%i]%mod;

int n;

struct P{

    int s,b,id;

    bool operator <(const P& c) const{

        if(s==c.s) return b>c.b;

        return s<c.s;

    }

}p[maxn];

int g[maxn],pre[maxn];

void init(){

    mt(g,);mt(pre,-);g[] = -;

}

void output(int x){

    if(pre[x]==-){

        printf("%d",p[x].id);

        return;

    }

    output(pre[x]);

    printf(" %d",p[x].id);

}

int main(){

    init();

    scanf("%d",&n);

    rep(i,,n+) scanf("%d%d",&p[i].s,&p[i].b),p[i].id = i;

    sort(p+,p+n+);

    int ans = ;

    rep(i,,n+){

        int l = , r = ans, pos = -;

        while(l <= r){ //在前面找出找到第一个大于等于当前的b的点

            int mid = (l + r) >> ;

            if(p[i].b > p[g[mid]].b) l = mid + ;

            else pos = mid,r = mid - ;

        }

        if(pos==-) ans++, pos = ans;

        pre[i] = g[pos - ];

        if(p[i].b < p[g[pos]].b || g[pos] == ) g[pos] = i;

    }

    printf("%d\n",ans);

    output(g[ans]);

    return ;

}

巴特西

今日SGU 5.20

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