中国大学生程序设计竞赛（秦皇岛）正式赛东北大学秦皇岛分校（SMU Autumn 2024 Team Round 1）
Problem A. 贵校是构造王国吗 I
思路
官方题解很清晰明了。
代码
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'
#define PII pair<int,int>
const int N=1e6+3;
void solve() {int n,k;cin>>n>>k;int dq=1;vector<PII>ans;set<PII>st;ans.push_back({1,1});st.insert({1,1});dq++;int ls=1;for (int i = 2; i <=n ; ++i) {ans.push_back({i,ls});st.insert({i,ls});dq++;ls++;ans.push_back({i,ls});st.insert({i,ls});dq++;}ans.push_back({1,n});st.insert({1,n});dq++;if(dq>k){for(auto [x,y]:ans){cout<<x<<' '<<y<<endl;}return ;}else{for (int i = 1; i <=n ; ++i) {for (int j = 1; j <=n ; ++j) {if(st.count({i,j})==0){ans.push_back({i,j});if(ans.size()==k){for(auto [x,y]:ans){cout<<x<<' '<<y<<endl;}return ;}}}}}}
signed main()
{ios::sync_with_stdio(false),cin.tie(0);int t=1;// cin>>t;while(t--){solve();}return 0;
}
Problem D. 茶和咖啡
思路
和题解类似。不过我们是用线段树维护前缀最小的物品。
代码
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define PII pair<int,int>
#define INF INT_MAX
#define N 200010
#define lc p<<1
#define rc p<<1|1
int n, w[N];
int a[N],b[N];
struct node{int l,r;PII mi;
}tr[N<<2],tr1[N<<2];
void push_up(int p){tr[p].mi=min(tr[lc].mi,tr[rc].mi);
}
void build(int p,int l,int r){tr[p]={l,r,{0,0}};if(l==r){tr[p]={l,r,{w[l]-b[l],l}};return;}int mid=(l+r)>>1;build(lc,l,mid);build(rc,mid+1,r);push_up(p);
}
void update(int p,int x,int k){if(tr[p].l==x and tr[p].r==x){tr[p].mi.first+=k;return;}int mid=tr[p].l+tr[p].r>>1;if(x<=mid)update(lc,x,k);if(x>mid)update(rc,x,k);push_up(p);
}
PII query(int p,int x,int y){if(x<=tr[p].l and tr[p].r<=y){return tr[p].mi;}int mid=tr[p].l+tr[p].r>>1;PII sum={INF,0};if(x<=mid){sum=min(sum, query(lc,x,y));}if(y>mid){sum=min( query(rc,x,y),sum);}return sum;
}void push_up1(int p){tr1[p].mi=min(tr1[lc].mi,tr1[rc].mi);
}
void build1(int p,int l,int r){tr1[p]={l,r,{0,0}};if(l==r){tr1[p]={l,r,{w[l]-b[l],l}};return;}int mid=(l+r)>>1;build1(lc,l,mid);build1(rc,mid+1,r);push_up1(p);
}
void update1(int p,int x,int k){if(tr1[p].l==x and tr1[p].r==x){tr1[p].mi.first+=k;return;}int mid=tr1[p].l+tr1[p].r>>1;if(x<=mid)update1(lc,x,k);if(x>mid)update1(rc,x,k);push_up1(p);
}
PII query1(int p,int x,int y){if(y==0)return {INF,0};if(x<=tr1[p].l and tr1[p].r<=y){return tr1[p].mi;}int mid=tr1[p].l+tr1[p].r>>1;PII sum={INT_MAX,0};if(x<=mid){sum=min(sum, query1(lc,x,y));}if(y>mid){sum=min( query1(rc,x,y),sum);}return sum;
}void solve(){int q;cin>>n>>q;for (int i = 1; i <=n ; ++i) {cin>>w[i];a[i]=b[i]=0;}build(1,1,n);for(int i=1;i<=q;i++){int x,y;cin>>x>>y;a[x]+=y;}b[n+1]=0;for(int i=n;i>=1;i--){b[i]=a[i]+b[i+1];}build1(1,1,n);int pos=n,s=0;int ans=0;for(int i=1;i<=n;i++){auto[mn,pp]=query1(1,1,pos);mn+=s;if(pos<n){auto[mn1,pp1]=query(1,pos+1,n);if(mn1<mn){mn=mn1;pp=pp1;}}if(pp<=pos){pos=pp-1;s=b[pp];} update(1,pp,INF);update1(1,pp,INF);ans+=mn;cout<<ans<<' ';}cout<<endl;
}signed main(){ios::sync_with_stdio(false);cin.tie(0);int T=1;cin>>T;while(T--){solve();}return 0;
}Problem G. 最大路径
思路
懒得写了，题解写得很清楚，赞
代码
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'
#define PII pair<int,int>
const int N=1e6+3;
void solve() {int n,m;cin>>n>>m;int ans=0;vector<int>a(n),b(m);for (int i = 0; i <n ; ++i) {cin>>a[i];}for (int i = 0; i <m ; ++i) {cin>>b[i];}for (int i = 1; i <n ; ++i) {ans+=abs(a[i-1]-a[i]);}for (int i = 1; i <m ; ++i) {ans+=abs(b[i-1]-b[i]);}cout<<ans<<endl;}
signed main()
{ios::sync_with_stdio(false),cin.tie(0);int t=1;// cin>>t;while(t--){solve();}return 0;
}
Problem J. 维克多词典
思路
子集 DP。
单词长度很小，可以考虑状压DP求答案。
设 \(dp_S\) 为为学习完 S 的集合长度的单词的最小天数，转移的时候枚举每个 S 的子集即可。
代码
#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
#define int long long
#define double long double
#define LL __int128
#define PII pair<double,double>
#define PIC pair<int,char>
#define endl '\n'
typedef long long ll;
const int N = 1e4 + 10;
int a[100];
vector<int>q;
PII S[N];
vector<int>g[20];
void solve() {int n, w;cin >> n >> w;int t = 0;for (int i = 1; i <= n; i++) {int x;cin >> x;if (!a[x]) {t++;q.push_back(x);}a[x]++;}for(int i=0;i<(1<<t);i++){S[i]={1e9,1e9};g[__builtin_popcount(i)].push_back(i);}S[0]={0,0};for (int i = 0; i < t; ++i){for(auto j:g[i]){for(int k=0;k<t;k++){if ((j >> k & 1)==0) {PII x=S[j];if(a[q[k]]+x.second<=w){x.second+=a[q[k]];}else{x.first++;x.second=a[q[k]];}S[j^(1<<k)]=min(S[j^(1<<k)],x);}}}}cout<<S[(1<<t)-1].first+(S[(1<<t)-1].second>0)<<endl;}
signed main() {ios::sync_with_stdio(false);cin.tie(0);int t = 1;
// cin>>t;while (t--) {solve();}return 0;
}
