Codeforces Global Round 3(A-D)

我凉了。。感觉自己啥都不会做，搞不好起床就绿了啊

代码和反思起床补，今天要反了个大思的

A. Another One Bites The Dust

把所有的ab排在一起然后两边叉a和b

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<stdlib.h>

#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
#define lowbit(x) ((x)&(-x)) 

using namespace std;

typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;

const db eps = 1e-;
const int mod = 1e9+;
const int maxn = 2e6+;
const int maxm = 2e6+;
const int inf = 0x3f3f3f3f;
const db pi = acos(-1.0);

int main(){
    ll a, b,c;
    scanf("%lld %lld %lld", &a, &b, &c);
    ll ans = c*2ll;
    ll tmp = min(a,b);
    ans+=tmp*;
    a-=tmp;b-=tmp;
    if(a>||b>)ans++;
    printf("%lld",ans);
    return ;
}

B. Born This Way

题意：有n个从A到B的航班，m个从B到C的航班，要你删除k个使得到达C的最晚，求最晚时间

思路：删除的方式一定是前i个从A到B的，和k-i个最早从A到B之后能乘坐的B到C的航班。由于对于每个i，这个方案是惟一的，枚举即可。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<stdlib.h>

#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
#define lowbit(x) ((x)&(-x)) 

using namespace std;

typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;

const db eps = 1e-;
const int mod = 1e9+;
const int maxn = 3e6+;
const int maxm = 2e6+;
const int inf = 0x3f3f3f3f;
const db pi = acos(-1.0);

ll a[maxn],b[maxn];
ll n,m,ta,tb,k;
ll mi[maxn],mx[maxn];
int main(){
    scanf("%lld %lld %lld %lld %lld",&n, &m,&ta,&tb,&k);
    for(int i = ; i <= n; i++){
        scanf("%lld", &a[i]);a[i]+=ta;
    }
    for(int i = ; i <= m; i++){
        scanf("%lld", &b[i]);
    }ll res = -;
    if(k>=n||k>=m)return printf("-1"),;
    for(int i = ; i <= k; i++){
        int t = k-i;
        int p = lower_bound(b+,b++m,a[i+])-b;
        p+=t;
        if(p>m)return printf("-1"),;
        else res = max(res, b[p]+tb);
    }printf("%lld", res);
    return ;
}

C. Crazy Diamond

题意：给你一个1-n的排列，每次选择两个位置i,j交换位置上的值，且满足2*|i-j|>=n，让你输出具体方案，使得operation不超过5n。

思路：根据限制条件，发现位置1和n合起来可以调动整个数组，而n/2和n/2+1位置只能分别和n和1交换，所以可以从中间开始向两边分别填充，操作不超过3n

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>

#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
#define lowbit(x) ((x)&(-x)) 

using namespace std;

typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;

const db eps = 1e-;
const int mod = 1e9+;
const int maxn = 2e6+;
const int maxm = 2e6+;
const int inf = 0x3f3f3f3f;
const db pi = acos(-1.0);

int n;
int a[maxn];
int id[maxn];
vector<PI>v;
void swp(int x, int y){
    if(x==y)return;
    int i = a[x];
    int j = a[y];
    swap(a[x],a[y]);
    swap(id[i],id[j]);
    v.pb({x,y});
}
int main() {
    scanf("%d", &n);
    for(int i = ; i <= n; i++){
        scanf("%d", &a[i]);
        id[a[i]]=i;
    }
    for(int i = n/; i >= ; i--){
        int j = n-i+;
        if(id[i]!=i){
            if(id[i]<n/){
                swp(n,id[i]);
            }
            else{
                swp(,id[i]);
                swp(n,);
            }
            swp(id[i],i);
        }

        if(id[j]!=j){
            if(id[j]>n/){
                swp(,id[j]);
            }
            else{
                swp(n,id[j]);
                swp(n,);
            }
            swp(id[j],j);
        }
    }
    //for(int i = 1; i <= n; i++)printf("%d ",a[i]);printf("\n");
    printf("%d\n",v.size());
    for(int i = ; i < (int)v.size(); i++){
        printf("%d %d\n",v[i].fst,v[i].sc);
    }
    return ;
}

D. Dirty Deeds Done Dirt Cheap

题意：给你n个pair，让你取出其中一些排列，使得他们排成一行的数x1>x2<x3>x4<x5...或x1<x2>x3<x4>x5...，问这样排列最长的方案

思路：我好弱智啊。。对于pair内不等号方向相同的分别考虑，比如在每个pair都是x<y的时候，对ans中两个相邻的pair，都有x1<y1>x2<y2。因为y1>x1，所以我们让x1>x2就可以满足y1>x2，所以对该类所有pair排序即可，答案为倒序输出。。。同样地，对x>y的时候只需要排序后正序输出。。把这两种size大的输出即可

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<stdlib.h>

#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
#define lowbit(x) ((x)&(-x)) 

using namespace std;

typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;

const db eps = 1e-;
const int mod = 1e9+;
const int maxn = 3e6+;
const int maxm = 2e6+;
const int inf = 0x3f3f3f3f;
const db pi = acos(-1.0);

struct Node{
    int x, y;
    int id;
}node[maxn];vector<Node>a,b;
bool cmp(Node a, Node b){
    if(a.x==b.x)return a.y<b.y;
    return a.x<b.x;
}
int main(){
    int n;
    scanf("%d", &n);
    for(int i = ; i <= n; i++){
        node[i].id=i;
        scanf("%d %d" ,&node[i].x, &node[i].y);
        if(node[i].x<node[i].y)a.pb(node[i]);
        else b.pb(node[i]);
    }
    printf("%d\n",max(a.size(), b.size()));
    sort(a.begin(),a.end(),cmp);
    sort(b.begin(),b.end(),cmp);
    if(a.size()>b.size()){
        for(int i = a.size()-; i>=; i--){
            printf("%d ",a[i].id);
        }
    }
    else{
        for(int i = ; i < b.size(); i++)printf("%d ",b[i].id);
    }
    return ;
}