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;
}

赛后补题

G. The Last Cumulonimbus Cloud（拓扑排序好题）

任意非空子图都有一个度不超过10的点=可以把度>10的点的贡献都归到这些不超过10的点上

拓扑排序每删掉一个度数不足10的点就会多出一个不足10的点

最后图可以化成一个联通且每个点出度均不超过10的dag

u加的时候把贡献推到u的下游

u算的时候上游已经推给过u，只需要再加上u的所有下游的贡献