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 ;
}