ICPC2024 邀请赛西安站（7/8/13）

心得

[ICPC2024 Xi'an I] ICPC2024 邀请赛西安站重现赛 - 比赛详情 - 洛谷

7表示赛时ac了7个，8表示含补题总共ac数，13表示题目总数

题目

M. Chained Lights

打表，发现只有k=1是YES

//#include <bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<vector>
#include<map>
using namespace std;
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define per(i,a,b) for(int i=(a);i>=(b);--i)
typedef long long ll;
typedef double db;
typedef pair<int,int> P;
#define fi first
#define se second
#define pb push_back
#define dbg(x) cerr<<(#x)<<":"<<x<<" ";
#define dbg2(x) cerr<<(#x)<<":"<<x<<endl;
#define SZ(a) (int)(a.size())
#define sci(a) scanf("%d",&(a))
#define pt(a) printf("%d",a);
#define pte(a) printf("%d\n",a)
#define ptlle(a) printf("%lld\n",a)
#define debug(...) fprintf(stderr, __VA_ARGS__)
using namespace std;
const int N=1e6+10;
int t,n,k,fac[N],sum[N],light[N];
void press(int x)
{
    light[x]^=1;
    for (int y=x+x;y<=n;y+=x) press(y);
}
int main(){
    sci(t);
    while(t--){
        sci(n),sci(k);
        puts(k==1?"YES":"NO");
    }
    return 0;
}

J. Triangle

数三角形，手玩发现一些规律，

比如：n=3，m=9实际是15，然后发现和gcd有关

//#include <bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<vector>
#include<map>
using namespace std;
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define per(i,a,b) for(int i=(a);i>=(b);--i)
typedef long long ll;
typedef double db;
typedef pair<int,int> P;
#define fi first
#define se second
#define pb push_back
#define dbg(x) cerr<<(#x)<<":"<<x<<" ";
#define dbg2(x) cerr<<(#x)<<":"<<x<<endl;
#define SZ(a) (int)(a.size())
#define sci(a) scanf("%d",&(a))
#define pt(a) printf("%d",a);
#define pte(a) printf("%d\n",a)
#define ptlle(a) printf("%lld\n",a)
#define debug(...) fprintf(stderr, __VA_ARGS__)
using namespace std;
const int N=1e6+10;
int a,b;
ll gcd(ll a,ll b){
    return !b?a:gcd(b,a%b);
}
int main(){
    sci(a),sci(b);
    // if(a==b){
    //     printf("%lld\n",1ll*a*(a+1)/2);
    // }
    // else{
    ll v=gcd(a,b);
    ptlle((1ll*a*b-1ll*v*(a/v)*(b/v))/2+1ll*((a/v)*(b/v)+1)/2*v);
    //}
    return 0;
}
/*
3 9 = 15
*/

D. Make Them Straight

枚举k，根据ai-i*k确定能在同一个序列里的子序列，子序列里的是不改的

首项得是正的，后面i*k>1e6说明一定要改

剪枝一下复杂度是可接受的O(klogn)

//#include <bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<vector>
#include<map>
using namespace std;
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define per(i,a,b) for(int i=(a);i>=(b);--i)
typedef long long ll;
typedef double db;
typedef pair<int,int> P;
#define fi first
#define se second
#define pb push_back
#define dbg(x) cerr<<(#x)<<":"<<x<<" ";
#define dbg2(x) cerr<<(#x)<<":"<<x<<endl;
#define SZ(a) (int)(a.size())
#define sci(a) scanf("%d",&(a))
#define pt(a) printf("%d",a);
#define pte(a) printf("%d\n",a)
#define ptlle(a) printf("%lld\n",a)
#define debug(...) fprintf(stderr, __VA_ARGS__)
using namespace std;
const int N=2e5+10;
int n,a[N],b[N];
map<ll,ll>mp;
ll sum,ans;
int main(){
    sci(n);
    rep(i,0,n-1)sci(a[i]);
    rep(i,0,n-1){
        sci(b[i]);
        sum+=b[i];
    }
    ans=sum;
    rep(k,0,1000000){
        mp.clear();
        ll res=0;
        rep(i,0,n-1){
            ll v=a[i]-1ll*i*k;
            if(1ll*i*k>1000000)break;
            if(v<0)continue;
            mp[v]+=b[i];
            res=max(res,mp[v]);
            //if(mp[v]==9)printf("i:%d v:%lld mp:%lld\n",i,v,mp[v]);
        }
        ans=min(ans,sum-res);
    }
    ptlle(ans);
    return 0;
}
/*
3 9 = 15
*/

L. Rubbish Sorting

二进制子集枚举一下，大概sosdp的思想，因为|s|<=5

#include <bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<vector>
#include<map>
using namespace std;
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define per(i,a,b) for(int i=(a);i>=(b);--i)
typedef long long ll;
typedef double db;
typedef pair<int,int> P;
#define fi first
#define se second
#define pb push_back
#define dbg(x) cerr<<(#x)<<":"<<x<<" ";
#define dbg2(x) cerr<<(#x)<<":"<<x<<endl;
#define SZ(a) (int)(a.size())
#define sci(a) scanf("%d",&(a))
#define pt(a) printf("%d",a);
#define pte(a) printf("%d\n",a)
#define ptlle(a) printf("%lld\n",a)
#define debug(...) fprintf(stderr, __VA_ARGS__)
using namespace std;
const int N=2e5+10,M=2e7+10,INF=0x3f3f3f3f;
int q,n,op,x,y,bs[8];
char s[N];
int val[M],now=INF;
int main(){
    memset(val,INF,sizeof val);
    bs[0]=1;
    rep(i,1,6)bs[i]=bs[i-1]*28;
    sci(q);
    while(q--){
        sci(op);
        scanf("%s",s);
        n=strlen(s);
        if(op==1){
            sci(y);
            int w=0;
            rep(i,0,n-1){
                int v=s[i]-'a'+1;
                w+=v*bs[i];
            }
            val[w]=min(val[w],y);
            now=min(now,y);
            rep(p,1,n){
                int up=(1<<p)-1;
                rep(i,0,up-1){
                    int w=0;
                    rep(j,0,p-1){
                        int z=i>>j&1,v=0;
                        if(z)v=27;
                        else v=(s[j]-'a'+1);
                        w+=v*bs[j];
                    }
                    val[w]=min(val[w],y);
                }
            }
        }
        else{
            int w=0;
            rep(i,0,n-1){
                int v=s[i]-'a'+1;
                w+=v*bs[i];
            }
            if(val[w]!=INF){
                pte(val[w]);
                continue;
            }
            vector<int>mn(n+1,INF);
            mn[n]=now;
            rep(p,1,n){
                int up=(1<<p)-1;
                rep(i,0,up-1){
                    int w=0,tot=0;
                    rep(j,0,p-1){
                        int z=i>>j&1,v=0;
                        if(z)v=27,tot++;
                        else v=(s[j]-'a'+1);
                        w+=v*bs[j];
                    }
                    mn[tot+n-p]=min(mn[tot+n-p],val[w]);
                }
            }
            rep(i,0,n){
                if(mn[i]!=INF){
                    //printf("i:%d mn:%d\n",i,mn[i]);
                    pte(mn[i]);
                    break;
                }
            }
        }
    }
    return 0;
}
/*
3 9 = 15
*/

B. Turn Off The Lights（构造）

最终一定是和某一列完全相同/完全相反的，

不妨和第i列完全相同，全是01011，那么再把这三列的1按列取消掉就可以了

所以枚举和哪列相同，bitset加速统计，复杂度O(n^3/64)

#include <bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<vector>
#include<map>
using namespace std;
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define per(i,a,b) for(int i=(a);i>=(b);--i)
typedef long long ll;
typedef double db;
typedef pair<int,int> P;
#define fi first
#define se second
#define pb push_back
#define dbg(x) cerr<<(#x)<<":"<<x<<" ";
#define dbg2(x) cerr<<(#x)<<":"<<x<<endl;
#define SZ(a) (int)(a.size())
#define sci(a) scanf("%d",&(a))
#define pt(a) printf("%d",a);
#define pte(a) printf("%d\n",a)
#define ptlle(a) printf("%lld\n",a)
#define debug(...) fprintf(stderr, __VA_ARGS__)
using namespace std;
const int N=1e3+10;
int n,k,a[N][N],sum[N];
bitset<1005>b[2][N];
bool flip[N];
bool ck(int i){
    int sum=0;
    rep(j,1,n){
        if(j==i)continue;
        int s1=(b[1][i]^b[0][j]).count(),s2=(b[0][i]^b[0][j]).count();
        if(s1<s2)flip[j]=1;
        else flip[j]=0;
        sum+=min(s1,s2);
    }
    return sum<=k;
}
void out(int i){
    //printf("i:%d\n",i);
    vector<P>ans;
    rep(j,1,n){
        if(j==i)continue;
        if(flip[j])ans.pb(P(j,0));
        rep(x,1,n){
            if(flip[j]){
                if(a[j][x]!=(a[i][x]^1))ans.pb(P(j,x));
            }
            else{
                if(a[j][x]!=a[i][x])ans.pb(P(j,x));
            }
        }
    }
    rep(x,1,n){
        if(a[i][x])ans.pb(P(0,x));
    }
    pte(SZ(ans));
    for(auto x:ans){
        printf("%d %d\n",x.fi,x.se);
    }
}
int main(){
    sci(n),sci(k);
    rep(i,1,n){
        rep(j,1,n){
            sci(a[i][j]);
            if(a[i][j])b[0][i].set(j);
            else b[1][i].set(j);
        }
    }
    rep(i,1,n){
        rep(x,0,1){
            if(ck(i)){
                out(i);
                return 0;
            }
        }
    }
    puts("-1");
    return 0;
}
/*
3 9 = 15
*/

F. XOR Game（博弈）

先从大到小考虑ai=1的值，z是0的个数算最小的数

因为操作一次就可以获得收益/ban掉对应收益，所以alice和bob会先抢这部分

剩下的局面，尽可能避免2的出现，全是2的情况，谁先操作谁就输了，

所以判一下剩下局面的奇偶性，奇数alice全取，偶数bob全ban

#include <bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<vector>
#include<map>
using namespace std;
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define per(i,a,b) for(int i=(a);i>=(b);--i)
typedef long long ll;
typedef double db;
typedef pair<int,int> P;
#define fi first
#define se second
#define pb push_back
#define dbg(x) cerr<<(#x)<<":"<<x<<" ";
#define dbg2(x) cerr<<(#x)<<":"<<x<<endl;
#define SZ(a) (int)(a.size())
#define sci(a) scanf("%d",&(a))
#define pt(a) printf("%d",a);
#define pte(a) printf("%d\n",a)
#define ptlle(a) printf("%lld\n",a)
#define debug(...) fprintf(stderr, __VA_ARGS__)
using namespace std;
const int N=1e5+10;
int k,z,a[N],b[N],c;
int main(){
    sci(k);sci(z);
    rep(i,0,k-1){
        sci(a[i]);
    }
    per(i,k-1,0){
        if(a[i]==1){
            c++;
            if(c&1)b[i]=1;
            a[i]=0;
        }
    }
    per(i,k-1,0){
        if(a[i]&1)c++;
    }
    c+=(z&1);
    per(i,k-1,0){
        if(!a[i])continue;
        if(c&1)b[i]=1;
    }
    per(i,k-1,0){
        printf("%1d",b[i]);
    }
    puts("");
    return 0;
}
/*
3 9 = 15
*/

I. Smart Quality Inspector（状压dp+区间dp）

避免后效性，肯定是从大到小考虑max值，

被前面的最大值覆盖了的区间，再选最大值时就没有贡献了

dp[S]表示当前最大值已经覆盖的状态是S时的最大代价和

b[i][l][r]表示区间包含第i位且被[l,r]完整包含的区间的数量

新增一位时，往左往右拓展0的区域找到最左最右，这一段的b值对应的区间都是有贡献的

复杂度O(n^5+n*2^n)

#include <bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<vector>
#include<map>
using namespace std;
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define per(i,a,b) for(int i=(a);i>=(b);--i)
typedef long long ll;
typedef double db;
typedef pair<int,int> P;
#define fi first
#define se second
#define pb push_back
#define dbg(x) cerr<<(#x)<<":"<<x<<" ";
#define dbg2(x) cerr<<(#x)<<":"<<x<<endl;
#define SZ(a) (int)(a.size())
#define sci(a) scanf("%d",&(a))
#define pt(a) printf("%d",a);
#define pte(a) printf("%d\n",a)
#define ptlle(a) printf("%lld\n",a)
#define debug(...) fprintf(stderr, __VA_ARGS__)
using namespace std;
const int N=25,M=(1<<20)+5,INF=0x3f3f3f3f;
int n,k,m,l,r,a[N][N],b[N][N][N],dp[M];
void upd(int &x,int y){
    x=min(x,y);
}
int main(){
    sci(n),sci(k),sci(m);
    rep(i,1,m){
        sci(l),sci(r);
        a[l][r]++;
    }
    rep(i,1,n){
        rep(l,1,i){
            rep(r,i,n){
                rep(x,l,i){
                    rep(y,i,r){
                        b[i][l][r]+=a[x][y];
                    }
                }
            }
        }
    }
    memset(dp,INF,sizeof dp);
    dp[0]=0;
    int up=(1<<n)-1;
    rep(i,0,up){
        vector<int>pre(n,-1),suf(n,n);
        int now=-1,bit=0;
        rep(j,0,n-1){
            int v=i>>j&1;
            if(v==1)now=j,bit++;
            else pre[j]=now+1;
        }
        now=n;
        per(j,n-1,0){
            int v=i>>j&1;
            if(v==1)now=j;
            else{
                suf[j]=now-1;
                int x=j+1,y=pre[j]+1,z=suf[j]+1;
                upd(dp[i|(1<<j)],dp[i]+b[x][y][z]*max(k-bit,0));
            }
        }
    }
    pte(dp[up]);
    return 0;
}