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-6; const int mod = 1e9+7; const int maxn = 2e6+100; const int maxm = 2e6+100; 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*2; a-=tmp;b-=tmp; if(a>0||b>0)ans++; printf("%lld",ans); return 0; }
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-6; const int mod = 1e9+7; const int maxn = 3e6+100; const int maxm = 2e6+100; 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 = 1; i <= n; i++){ scanf("%lld", &a[i]);a[i]+=ta; } for(int i = 1; i <= m; i++){ scanf("%lld", &b[i]); }ll res = -1; if(k>=n||k>=m)return printf("-1"),0; for(int i = 0; i <= k; i++){ int t = k-i; int p = lower_bound(b+1,b+1+m,a[i+1])-b; p+=t; if(p>m)return printf("-1"),0; else res = max(res, b[p]+tb); }printf("%lld", res); return 0; }
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-6; const int mod = 1e9+7; const int maxn = 2e6+100; const int maxm = 2e6+100; 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 = 1; i <= n; i++){ scanf("%d", &a[i]); id[a[i]]=i; } for(int i = n/2; i >= 1; i--){ int j = n-i+1; if(id[i]!=i){ if(id[i]<n/2){ swp(n,id[i]); } else{ swp(1,id[i]); swp(n,1); } swp(id[i],i); } if(id[j]!=j){ if(id[j]>n/2){ swp(1,id[j]); } else{ swp(n,id[j]); swp(n,1); } swp(id[j],j); } } //for(int i = 1; i <= n; i++)printf("%d ",a[i]);printf(" "); printf("%d ",v.size()); for(int i = 0; i < (int)v.size(); i++){ printf("%d %d ",v[i].fst,v[i].sc); } return 0; }
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-6; const int mod = 1e9+7; const int maxn = 3e6+100; const int maxm = 2e6+100; 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 = 1; 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 ",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()-1; i>=0; i--){ printf("%d ",a[i].id); } } else{ for(int i = 0; i < b.size(); i++)printf("%d ",b[i].id); } return 0; }