The 2025 Sichuan Provincial Collegiate Programming Contest 复盘

初看没啥思路，但是这个数据范围出奇的小；

其实这道题让我们觉得困难的原因是因为它有两个变量，如果变成仅对一个变量求最小值，这其实就是一个简单的完全背包问题；

因为M<=1000,val<=200,所以最终每个元素和的可能取值不会超过200 000，

考虑对于每一种存在的a的和的取值，求出它的b的和的最小值，然后对二者乘积找最小值；

code:

#include<bits/stdc++.h>

#define int long long

#define inf 0x3f3f3f3f3f3f3f

#define GG ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

#define cnot cout<<"NO"<<"\n"

#define cyes cout<<"YES"<<"\n"

#define cans cout<<ans<<"\n"

#define pb push_back

#define x0 first

#define y0 second

#define lc p<<1

#define rc p<<1|1

#define mem(a,b) memset(a,b,sizeof(a))

#define sp(x) fixed<<setprecision(x)

#define all(v) v.begin(),v.end()

#define fr(i,st,ed) for(int i=st;i<=ed;i++)

#define ffr(i,st,ed,dt) for(int i=st;i<=ed;i+=dt)

using namespace std;

typedef pair<int,string>Pis;

typedef pair<int,int>Pii;

typedef pair<string,string>Pss;

//const int N=1000*200+5,mod=1e9+7,M=1e6+10;

int lowbit(int x){

return x&(-x);}

void solve(){

int n,m;

cin>>n>>m;

vector<int>u(m+1),v(m+1),a(m+1),b(m+1);

fr(i,1,m){

cin>>u $i$ >>v $i$ >>a $i$ >>b $i$ ;

}

//dp $point$ $costa$ =mincostb

int N=200*m+5;

vector<vector<int>>dp(n+1,vector<int>(N,inf));

dp $1$ $0$ =0;

fr(C,1,N-1){

fr(i,1,m){

if(C>=a $i$ ){

dp $v\[i$ ] $C$ =min(dp $v\[i$ ] $C$ ,dp $u\[i$ ] $C-a\[i$ ]+b $i$ );

}

int minx=inf;

int ans1=0;

int ans2=0;

fr(i,1,N-1){

if(dp $n$ $i$ !=inf){

if(dp $n$ $i$ *i<minx){

minx=dp $n$ $i$ *i;

ans1=i;

ans2=dp $n$ $i$ ;

}

cout<<ans1<<" "<<ans2<<"\n";

}

signed main(){

GG;

int _t=1;

cin>>_t;

while(_t--){

solve();

}

2026蓝桥省B LQ序列原题

将每个问号同时看成1，0，统计1的前缀/0的后缀

对于每个？，比较前后缀大小，确定该点为0/1的贡献大

code:

#include<bits/stdc++.h>

#define ll long long

#define inf 0x3f3f3f3f3f3f3f

#define GG ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

#define cnot cout<<"NO"<<"\n"

#define cyes cout<<"YES"<<"\n"

#define cans cout<<ans<<"\n"

#define pb push_back

#define x0 first

#define y0 second

#define lc p<<1

#define rc p<<1|1

#define mem(a,b) memset(a,b,sizeof(a))

#define sp(x) fixed<<setprecision(x)

#define all(v) v.begin(),v.end()

#define fr(i,st,ed) for(int i=st;i<=ed;i++)

#define ffr(i,st,ed,dt) for(int i=st;i<=ed;i+=dt)

using namespace std;

typedef pair<int,string>Pis;

typedef pair<int,int>Pii;

typedef pair<string,string>Pss;

const int N=2e4+10,mod=1e9+7,M=1e6+10;

int lowbit(int x){

return x&(-x);}

void solve(){

int n;

cin>>n;

string s;

cin>>s;

s=" "+s;

vector<int>pl(n+5),eq(n+5),c(n+5),pre(n+5),ed(n+5);

for(int i=1;i<=n;i++){

if(s $i$ =='1'||s $i$ =='?')pl $i$ =pl $i-1$ +1;

else pl $i$ =pl $i-1$ ;

}

for(int i=n;i>=1;i--){

if(s $i$ =='0'||s $i$ =='?')eq $i$ =eq $i+1$ +1;

else eq $i$ =eq $i+1$ ;

}

for(int i=1;i<=n;i++){

if(s $i$ =='1')c $i$ =0;

else if(s $i$ =='0')c $i$ =1;

else{

if(pl $i$ <eq $i$ )c $i$ =0;

else c $i$ =1;

}

ll ans=0;

ll cnt=0;

for(int i=1;i<=n;i++){

if(c $i$ ==0)pre $i$ =pre $i-1$ +1;

else pre $i$ =pre $i-1$ ;

}

for(int i=1;i<=n;i++){

if(c $i$ ==1){

ans+=pre $i$ ;

}

cout<<ans<<"\n";

}

int main(){

GG;

int _t=1;

cin>>_t;

while(_t--){

solve();

}

注意下如果共线且距离<=2,需要两次移动来抵消共线方向的偏移；

code:

#include<bits/stdc++.h>

#define int long long

#define inf 0x3f3f3f3f3f3f3f

#define GG ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

#define cnot cout<<"NO"<<"\n"

#define cyes cout<<"YES"<<"\n"

#define cans cout<<ans<<"\n"

#define pb push_back

#define x0 first

#define y0 second

#define lc p<<1

#define rc p<<1|1

#define mem(a,b) memset(a,b,sizeof(a))

#define sp(x) fixed<<setprecision(x)

#define all(v) v.begin(),v.end()

#define fr(i,st,ed) for(int i=st;i<=ed;i++)

#define ffr(i,st,ed,dt) for(int i=st;i<=ed;i+=dt)

using namespace std;

typedef pair<int,string>Pis;

typedef pair<int,int>Pii;

typedef pair<string,string>Pss;

const int N=2e4+10,mod=1e9+7,M=1e6+10;

int lowbit(int x){

return x&(-x);}

void solve(){

int x,y,X,Y;

cin>>x>>y>>X>>Y;

int dt1=abs(X-x);

int dt2=abs(Y-y);

int dt=max(dt1,dt2);

if(dt1==0&&dt2==0){

cout<<0<<"\n";

return;

}

else if(dt1==0||dt2==0){

if(dt<=2){

cout<<2<<"\n";

return;

}

cout<<(dt+1)/2<<"\n";

}

signed main(){

GG;

int _t=1;

cin>>_t;

while(_t--){

solve();

}

字典树模板，反转插入看有多少节点即可

code:

#include<bits/stdc++.h>

#define int long long

#define inf 0x3f3f3f3f3f3f3f

#define GG ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

#define cnot cout<<"NO"<<"\n"

#define cyes cout<<"YES"<<"\n"

#define cans cout<<ans<<"\n"

#define pb push_back

#define x0 first

#define y0 second

#define lc p<<1

#define rc p<<1|1

#define mem(a,b) memset(a,b,sizeof(a))

#define sp(x) fixed<<setprecision(x)

#define all(v) v.begin(),v.end()

#define fr(i,st,ed) for(int i=st;i<=ed;i++)

#define ffr(i,st,ed,dt) for(int i=st;i<=ed;i+=dt)

using namespace std;

typedef pair<int,string>Pis;

typedef pair<int,int>Pii;

typedef pair<string,string>Pss;

const int N=3e5+10,mod=1e9+7,M=1e6+10;

int lowbit(int x){

return x&(-x);}

int idx=0;

int son $N$ $26$ ;

int cnt $N$ ;

int n;

void ist(string s){

int p=0;

int sz=s.size();

fr(i,0,sz-1){

int u=s $i$ -'a';

if(!son $p$ $u$ ){

son $p$ $u$ =++idx;

}

p=son $p$ $u$ ;

}

cnt $p$ ++;

}

void solve(){

cin>>n;

fr(i,1,n){

string s;

cin>>s;

reverse(all(s));

ist(s);

}

cout<<idx<<"\n";

}

signed main(){

GG;

int _t=1;

//cin>>_t;

while(_t--){

solve();

}

target string SCCPC;

我们不妨统计中间的C,看有多少条路径以它为中心

记录每一个C有多少个S与它相连；

记录每一个P有多少C与它相连；

那么对于每一个C, 搜索与它之间相邻的点，是c则统计相邻S的个数，是p则统计相邻c个数，-1（去掉自己）；

相乘

code:

#include<bits/stdc++.h>

#define int long long

#define inf 0x3f3f3f3f3f3f3f

#define GG ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

#define cnot cout<<"NO"<<"\n"

#define cyes cout<<"YES"<<"\n"

#define cans cout<<ans<<"\n"

#define pb push_back

#define x0 first

#define y0 second

#define lc p<<1

#define rc p<<1|1

#define mem(a,b) memset(a,b,sizeof(a))

#define sp(x) fixed<<setprecision(x)

#define all(v) v.begin(),v.end()

#define fr(i,st,ed) for(int i=st;i<=ed;i++)

#define ffr(i,st,ed,dt) for(int i=st;i<=ed;i+=dt)

using namespace std;

typedef pair<int,string>Pis;

typedef pair<int,int>Pii;

typedef pair<string,string>Pss;

const int N=2e4+10,mod=1e9+7,M=1e6+10;

int lowbit(int x){

return x&(-x);}

//tag:sccpc

// num1 c-s

// num2 p-c

//(s)c<-c->p(c)

void solve(){

int n;

cin>>n;

vector<char>c(n+1);

vector<int>num1(n+1),num2(n+1);

vector<vector<int>>g(n+1);

fr(i,1,n){

cin>>c $i$ ;

}

fr(i,1,n-1){

int u,v;

cin>>u>>v;

if(c $u$ =='C'){

if(c $v$ =='S'){

num1 $u$ ++;

}

if(c $v$ =='C'){

if(c $u$ =='S'){

num1 $v$ ++;

}

if(c $u$ =='P'){

if(c $v$ =='C'){

num2 $u$ ++;

}

if(c $v$ =='P'){

if(c $u$ =='C'){

num2 $v$ ++;

}

g $u$ .pb(v);

g $v$ .pb(u);

}

int ans=0;

fr(u,1,n){

int ans1=0;

int ans2=0;

if(c $u$ =='C'){

for(int v:g $u$ ){

if(c $v$ =='C'){

ans1+=num1 $v$ ;

}

if(c $v$ =='P'){

ans2+=(num2 $v$ -1);

}

ans=ans+ans1*ans2;

}

cans;

}

signed main(){

GG;

int _t=1;

cin>>_t;

while(_t--){

solve();

}

队友敲的

#include <bits/stdc++.h>

using namespace std;

using LL = long long;

#define endl "\n"

LL mod=998244353;

LL ksm(LL a,LL n){

LL res=1;

while(n){

if(n&1)res=res*a;

a=a*a;

n/=2;

}

return res;

}

vector<LL>fa(1e6),p(1e6),idx(1e6);

LL find(LL x){

if(x!=fa $x$ ){

fa $x$ =find(fa $x$ );

}

return fa $x$ ;

}

void union1(LL x,LL y){

LL a=find(x),b=find(y);

if(idx $a$ >idx $b$ )swap(a,b);

fa $a$ =fa $b$ ;

}

void solve(){

LL n;

cin>>n;

//vector<LL>p(n+1);

for(int i=1;i<=n;i++){

cin>>p $i$ ;

idx $p\[i$ ]=i;

fa $i$ =i;

}

vector<vector<LL>>edag(n+1);

for(int i=1;i<n;i++){

LL u,v;

cin>>u>>v;

edag $u$ .push_back(v);

edag $v$ .push_back(u);

}

vector<LL>ans(n+1);

for(int i=n;i>=1;i--){

LL x=p $i$ ;

for(int y:edag $x$ ){

if(idx $x$ <idx $y$ ){

LL a=find(y);

fa $a$ =x;

ans $a$ =x;

}

for(int i=1;i<=n;i++){

if(ans $i$ ==i)cout<<0<<" ";

else cout<<ans $i$ <<" ";

}

cout<<endl;

}

int main(){

ios::sync_with_stdio(0);

cin.tie(0),cout.tie(0);

LL T=1;

cin>>T;

while(T--){

solve();

}