Codeforces Round #703 (Div. 2) A-E 题解

Shifting Stacks
给你 n 长度的数组，每个下标包含 000-nnn 个木块其中可以将某一位置的木块向右边平移请问是否能凑出高度严格上升的数组一开始想简单了直接判断木块的和是否大于 (111+(n−1)(n-1)(n−1))*(n−1)(n-1)(n−1)/2 但是这样是错误的因为你不能一定凑出这样的排列假如都在最后一列所以需要按照符合情况所需最小木板数去做模拟 check

#include<cstdio>
#include<cstring>
#include<vector>
#include<cstring>
#include<queue>
#include<vector>
#include<map>
#include<set>
#include<iostream>
using namespace std;#define dbg(x) cout << #x << " = " << (x) <<endl;
#define dbg2(x,y) cout << #x << " = " << (x) << " " << #y << " = " << (y) <<endl;
#define dbg3(x,y,z) cout << #x << " = " << (x) << " " << #y << " = " << (y) <<" " << #z << " = " << z <<endl;
#define ll long long
#define fi first
#define se second
#define pb push_backconst int MAX_N = 105;
ll arr[MAX_N];int main()
{int t;scanf("%d",&t);while(t--){int n,x;ll sum = 0;scanf("%d",&n);int ck = n*(n-1)/2;for(int i = 1;i<=n;++i){scanf("%lld",&arr[i]);}bool flag = true;for(int i = 1;i<=n;++i){if(sum+arr[i]<(i-1)){flag = false;break;}sum+=arr[i]-(i-1);}if(flag) printf("YES\n");else printf("NO\n");}return 0;
}

Eastern Exhibition
给你 n 个点，横纵坐标的范围是 0−1e90-1e90−1e9 其实这个答案是会有很多种的，想到找中点，但是一开始nc写了个dfs(zz) 其实我们考虑假设nnn个点都在同一纵坐标，那么假如是偶数个点，在中间两个点(下标不同)之间任意平移不影响对xxx的贡献奇数点则不满足条件只能按着不动分类乘法一下即可 dfs可太秀了

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <iostream>
#include <set>
using namespace std;#define debug(x) cout << #x << " = " << (x) <<endl;
#define ll long long
const int MAX_N = 200025;ll arr[MAX_N],brr[MAX_N];int main()
{int t;scanf("%d",&t);while(t--){int n;scanf("%d",&n);for(int i = 1;i<=n;++i){scanf("%lld",&arr[i]);scanf("%lld",&brr[i]);}sort(arr+1,arr+1+n);sort(brr+1,brr+1+n);if(n%2==0){printf("%lld\n",(arr[n/2+1]-arr[n/2]+1)*(brr[n/2+1]-brr[n/2]+1));}else{printf("1\n");}}return 0;
}

Guessing the Greatest
比较有意思的交互题给你nnn个值不同的数组成的数组你可以选择一个区间进行询问，题目会告诉你这个区间内第二大的数的下标是什么要求你不超过20次查询得到最大数的下标所在地
222^202020 次方 > 1e61e61e6 所以我们自然想到二分
如何进行二分呢？
我们首先对整个数组查询一次，我们知道全局第二大的数在哪，那么我们可以每次判断一个区间是否含有最大值的check表达式为整个区间内第二大是否为全局第二大，那么经过一次check你就可以知道全局最大是在左区间还是在右区间（所在的区间查询得到的下标为一开始所求的全局第二大）那么就可以二分递归下去求解了

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<string>
using namespace std;#define dbg(x) cout << #x << " = "  << (x) << endl;int Find(int l,int r)
{int x;printf("? %d %d\n\n",l,r);fflush(stdout);scanf("%d",&x);return x;
}int main()
{bool flag = false;int n,ans;scanf("%d",&n);int xb = Find(1,n);if(xb<n&&Find(xb,n)==xb){int l = xb+1, r = n;while(l<=r){int mid = l+r>>1;int now_xb = Find(xb,mid);if(now_xb==xb){flag = true;ans = mid;r = mid - 1;}else{l = mid + 1;}}if(flag){printf("! %d\n",ans);fflush(stdout);}}if(xb>1&&!flag){int l = 1,r = xb-1;while(l<=r){int mid = l+r>>1;int now_xb = Find(mid,xb);if(now_xb==xb){flag = true;ans = mid;l = mid + 1;}else{r = mid - 1;}}if(flag){printf("! %d\n",ans);fflush(stdout);}}return 0;
}

Max Median
给你nnn个数组，mmm长度，你可以选择长度大于等于mmm的连续区间此区间正序排序 (n+1)/2(n+1)/2(n+1)/2向下取整的下标的数为 median 问你所有median中的最小值最大为多少
我们很自然的可以想到二分答案那么怎么将二分的答案与check过程联立呢？我们很自然的想到将原数大小大于等于二分所得答案的贡献为111，其余为−1-1−1 那么我们就可以去解决如何选出长度大于等于mmm的连续区间了
考虑前缀和如果两个前缀和值为一样，则代表他们之间的差的值为000，即中间这段区间可以选来我们发现区间值为000是没用的会取到下标为−1-1−1的数（与正序有关）所以我们只求前缀和 > 111的情况这样我们还得同时保证长度大于 mmm 那么不难想到维护一个前缀最小值代表前缀和的最小值然后利用前缀和求解即可

#include<cstdio>
#include<iostream>
#include<vector>
#include<map>
#include<algorithm>
#include<cstring>
#include<set>
#include<cmath>
#include<stack>
using namespace std;
const int MAX_N = 200025;
int arr[MAX_N],brr[MAX_N],sum[MAX_N],minn[MAX_N];
map<int,int> mp;
#define dbg(x) cout << #x << " = " << (x) << endl;
#define dbg2(x,y) cout << #x << " = " << (x) << " " << #y << " = " << (y) << endl;
#define dbg3(x,y,z) cout << #x << " = " << (x) << " " << #y << " = " << (y) << " " << #z << " = " << (z) << endl;int main()
{int n,m;scanf("%d%d",&n,&m);for(int i = 1;i<=n;++i){scanf("%d",&arr[i]);}minn[0] = 0x3f3f3f3f;sum[0] = 0;int l = 1, r = n;while(l<=r){bool flag = false;mp.clear();mp[0] = 0;int mid = l + r >>1;for(int i = 1;i<=n;++i){brr[i] = (arr[i]>=mid)?1:-1;sum[i] = sum[i-1]+brr[i];if(mp.find(sum[i])==mp.end()) mp[sum[i]] = i;minn[i] = min(minn[i-1],sum[i]);}for(int i = m;i<=n;++i){if(sum[i]>0) flag = true;if(i>m&&((sum[i]-minn[i-m])>0)) flag = true;}if(flag) l = mid + 1;else r = mid - 1;}printf("%d\n",r);return 0;
}

Paired Payment
最短路的两跳版本，由于题意比较明确，所以一上来就撸了个 n2n^2n2n2n^2n2 的最短路，这要是以前 zls 肯定说是肯定过不去的，然后我又会说我来几个玄学优化（每次都会被zls反驳哈哈，过去的时光还是很美好的），那么显然我一开始也是这样做了
后来去学习了题解，发现其实n2∗n2n^2*n^2n2∗n2的原因是因为我们想维护边但是图论里面有很重要的虚边的概念用虚点连接虚边我们得注意到这题的长度是505050（之前一直没有注意到）我们将每个点都乘 * 515151 代表虚点，每条边有两个状态，作为两跳中的第一跳，则与 v * 515151 + w 连一条边权为000的边，遍历上一跳边权从111-505050枚举作为第二条的边权 (was+w)(was+w) 后续用大跟堆最短路所求即可

#include<cstdio>
#include<iostream>
#include<vector>
#include<map>
#include<algorithm>
#include<cstring>
#include<set>
#include<cmath>
#include<stack>
#include<queue>
using namespace std;
const int MAX_N = 200025;
const int MAX_M = 200000*51+5;
const int INF = 1e9+7;#define ll long long
#define dbg(x) cout << #x << " = " << (x) << endl;
#define dbg2(x,y) cout << #x << " = " << (x) << " " << #y << " = " << (y) << endl;
#define dbg3(x,y,z) cout << #x << " = " << (x) << " " << #y << " = " << (y) << " " << #z << " = " << (z) << endl;vector<pair<int,int> > vt[MAX_M];
priority_queue<pair<int,int> > q;
int dp[MAX_M];void add(int u,int v,int w)
{vt[u*51].push_back(make_pair(v*51+w,0)); // middle pointfor(int was = 1;was<=50;++was){vt[u*51+was].push_back(make_pair(v*51,(was+w)*(was+w))); // end point}
}int main()
{int n,m,u,v,w;scanf("%d%d",&n,&m);for(int i = 1;i<=n*51;++i){dp[i] = INF;}for(int i = 1;i<=m;++i){scanf("%d%d%d",&u,&v,&w);u--,v--;add(u,v,w);add(v,u,w);}dp[0] = 0;q.push(make_pair(-dp[0],0));while(!q.empty()){pair<int,int> top = q.top();q.pop();for(auto v:vt[top.second]){if(dp[v.first]>-top.first+v.second){dp[v.first] = -top.first+v.second;q.push(make_pair(-dp[v.first],v.first));}}}for(int i=0;i<n;++i){if(dp[i*51]==INF) dp[i*51] = -1;i==n?printf("%d\n",dp[i*51]):printf("%d ",dp[i*51]);}return 0;
}

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