标签：col 水题 time show 字典 splay nod return isp
和队友一起打的两场cf之一，有些代码风格不（比）一(较)样(丑)
A. Beru-taxi
水题
#include<bits/stdc++.h> #define clr(a,b) memset(a,b,sizeof(a)) using namespace std; typedef long long ll; const int maxn=100010; int main(){ int n,zz; double a,b,z1,z2,z3,ans; while(~scanf("%lf %lf",&a,&b)) { scanf("%d",&n); zz = 1; for(int i = 0;i < n;i++) { scanf("%lf %lf %lf",&z1,&z2,&z3); if(zz) { zz = 0; ans = sqrt((z1 - a) * (z1 - a) + (z2 - b) * (z2 - b)) / z3; } else { ans = min(ans,sqrt((z1 - a) * (z1 - a) + (z2 - b) * (z2 - b)) / z3); } } printf("%.15f\n",ans); } }
B. Interesting drink
给出一堆数，q次询问，每次问比这个小的数有几个，sort加lower_bound()
#include<bits/stdc++.h> #define clr(a,b) memset(a,b,sizeof(a)) using namespace std; typedef long long ll; const int maxn=100010; vector<int >v; int n,q; int main(){ while(cin>>n) { v.clear(); for(int i=1;i<=n;i++) { int x; scanf("%d",&x); v.push_back(x); } sort(v.begin(),v.end()); cin>>q; while(q--) { int x; scanf("%d",&x); printf("%d\n",upper_bound(v.begin(),v.end(),x)-v.begin()); } } }
C. Hard problem
给出n个字符串，每个字符串反转需要$ci$的代价，问使得字符串在操作后按字典序非下降排的代价是多少。
$dp[i][0/1]$表示第$i$个串是否反转，然后在合法情况下的最小花费。
#include<bits/stdc++.h> #define clr(a,b) memset(a,b,sizeof(a)) using namespace std; typedef long long ll; const int maxn=100010; long long c[100010],dp[100010][3]; struct s { string s1,s2; }z[100010]; int main(){ int n,i,j,si; int flag; while(~scanf("%d",&n)) { for(i = 0;i < n;i++) { scanf("%lld",c + i); } for(i = 0;i < n;i++) { cin >> z[i].s1; si = z[i].s1.size(); for(j = si - 1;j >= 0;j--) { z[i].s2.push_back(z[i].s1[j]); } } //memset(dp,0,sizeof(dp)); dp[0][0] = 0; dp[0][1] = c[0]; flag = 1; for(i = 1;i < n;i++) { dp[i][0] = 1e18; dp[i][1] = 1e18; if(z[i].s1 >= z[i - 1].s1 && dp[i - 1][0] != 1e18) { dp[i][0] = min(dp[i][0],dp[i - 1][0]); } if(z[i].s1 >= z[i - 1].s2 && dp[i - 1][1] != 1e18) { dp[i][0] = min(dp[i][0],dp[i - 1][1]); } if(z[i].s2 >= z[i - 1].s1 && dp[i - 1][0] != 1e18) { dp[i][1] = min(dp[i][1],dp[i - 1][0] + c[i]); } if(z[i].s2 >= z[i - 1].s2 && dp[i - 1][1] != 1e18) { dp[i][1] = min(dp[i][1],dp[i - 1][1] + c[i]); } if(dp[i][0] == 1e18 && dp[i][1] == 1e18) { flag = 0; break; } } if(flag) { printf("%lld\n",min(dp[n - 1][0],dp[n - 1][1])); } else { printf("-1\n"); } } }
D. Vasiliy‘s Multiset
01字典树并且带删除操作，每次询问的时候从高位到低位询问，尽量往相反的地方走即可。
#include<bits/stdc++.h> #define clr(a,b) memset(a,b,sizeof(a)) using namespace std; typedef long long ll; const int maxn=200010; int q,x; char op[3]; int tr[maxn*31][2],tot=1; int vis[maxn*62]; void insert(int x) { int p=1; for(int k=30;k>=0;k--) { int ch=((x>>k)&1); if(tr[p][ch]==0)tr[p][ch]=++tot; p=tr[p][ch]; vis[p]++; } } void dele(int x){ int p=1; for(int k=30;k>=0;k--) { int ch=((x>>k)&1); p=tr[p][ch]; vis[p]--; } } int query(int x) { int res=0,p=1; for(int k=30;k>=0;k--) { int ch=((x>>k)&1); if(tr[p][!ch]&&vis[tr[p][!ch]]){ res|=(1<<k); p=tr[p][!ch]; }else{ p=tr[p][ch]; } } return res; } int main(){ while(cin>>q) { insert(0); while(q--) { scanf("%s %d",op,&x); if(op[0]==‘+‘){ insert(x); }else if(op[0]==‘-‘){ dele(x); }else{ printf("%d\n",query(x)); } } } }
E.Working routine
模拟一个二维的链表，每次交换就是改变一下链表的指针即可。
//#pragma comment(linker, "/STACK:102400000,102400000") #include<iostream> #include<stdio.h> #include<stdlib.h> #include<string.h> #include<string> #include<math.h> #include<cmath> #include<time.h> #include<map> #include<set> #include<vector> #include<queue> #include<algorithm> #include<numeric> #include<stack> #include<bitset> #include<unordered_map> const int maxn = 0x3f3f3f3f; const double EI = 2.71828182845904523536028747135266249775724709369995957496696762772407663035354594571382178525166427; const double PI = 3.141592653589793238462643383279; //#ifdef TRUETRUE //#define gets gets_s //#endif using namespace std; const int Maxn = 1010; struct node { int l, r, u, d, id, val; }a[Maxn * Maxn]; int n, m, q, cnt; int val[Maxn][Maxn]; int c[10][1010], d[10][1010],cc[10][1010],dd[10][1010]; int main(void) { //ios::sync_with_stdio(false); int i, j, A, B, C, D, H, W, pos1, pos2; while (cin >> n >> m >> q) { for (i = 0; i <= n + 1; i++) { for (j = 0; j <= m + 1; j++) { if (i == 0 || j == 0 || i == n + 1 || j == m + 1) { a[i * (m + 2) + j].val = 0; } else { scanf("%d", &a[i * (m + 2) + j].val); } a[i * (m + 2) + j].id = i * (m + 2) + j; } } for (i = 0; i <= n + 1; i++) { for (j = 0; j <= m + 1; j++) { if (i > 0) { a[i * (m + 2) + j].u = a[(i - 1) * (m + 2) + j].id; } if (i < n + 1) { a[i * (m + 2) + j].d = a[(i + 1) * (m + 2) + j].id; } if (j > 0) { a[i * (m + 2) + j].l = a[i * (m + 2) + j - 1].id; } if (j < m + 1) { a[i * (m + 2) + j].r = a[i * (m + 2) + j + 1].id; } } } while (q--) { int qq[10], qqq[10]; scanf("%d %d %d %d %d %d", &A, &B, &C, &D, &H, &W); pos1 = 0; for (i = 0; i < A; i++) { pos1 = a[a[pos1].d].id; } for (i = 0; i < B; i++) { pos1 = a[pos1].r; } pos2 = 0; for (i = 0; i < C; i++) { pos2 = a[pos2].d; } for (i = 0; i < D; i++) { pos2 = a[pos2].r; } qq[0] = pos1; int tmp1 = pos1; for (i = 0; i < W; i++) { c[0][i] = a[tmp1].u; cc[0][i] = tmp1; tmp1 = a[tmp1].r; } tmp1 = a[tmp1].l; qq[1] = tmp1; for (i = 0; i < H; i++) { c[1][i] = a[tmp1].r; cc[1][i] = tmp1; tmp1 = a[tmp1].d; } tmp1 = a[tmp1].u; qq[2] = tmp1; for (i = 0; i < W; i++) { c[2][i] = a[tmp1].d; cc[2][i] = tmp1; tmp1 = a[tmp1].l; } tmp1 = a[tmp1].r; qq[3] = tmp1; for (i = 0; i < H; i++) { c[3][i] = a[tmp1].l; cc[3][i] = tmp1; tmp1 = a[tmp1].u; } tmp1 = a[tmp1].d; qqq[0] = pos2; int tmp2 = pos2; for (i = 0; i < W; i++) { d[0][i] = a[tmp2].u; dd[0][i] = tmp2; tmp2 = a[tmp2].r; } tmp2 = a[tmp2].l; qqq[1] = tmp2; for (i = 0; i < H; i++) { d[1][i] = a[tmp2].r; dd[1][i] = tmp2; tmp2 = a[tmp2].d; } tmp2 = a[tmp2].u; qqq[2] = tmp2; for (i = 0; i < W; i++) { d[2][i] = a[tmp2].d; dd[2][i] = tmp2; tmp2 = a[tmp2].l; } tmp2 = a[tmp2].r; qqq[3] = tmp1; for (i = 0; i < H; i++) { d[3][i] = a[tmp2].l; dd[3][i] = tmp2; tmp2 = a[tmp2].u; } tmp2 = a[tmp2].d; int qq1[10], qq2[10]; memcpy(qq1, qq, sizeof(qq)); memcpy(qq2, qqq, sizeof(qqq)); for (i = 0; i < W; i++) { a[cc[0][i]].u = d[0][i]; a[d[0][i]].d = cc[0][i]; } for (i = 0; i < W; i++) { a[dd[0][i]].u = c[0][i]; a[c[0][i]].d = dd[0][i]; } for (i = 0; i < H; i++) { a[cc[1][i]].r = d[1][i]; a[d[1][i]].l = cc[1][i]; } for (i = 0; i < H; i++) { a[dd[1][i]].r = c[1][i]; a[c[1][i]].l = dd[1][i]; } for (i = 0; i < W; i++) { a[cc[2][i]].d = d[2][i]; a[d[2][i]].u = cc[2][i]; } for (i = 0; i < W; i++) { a[dd[2][i]].d = c[2][i]; a[c[2][i]].u = dd[2][i]; } for (i = 0; i < H; i++) { a[cc[3][i]].l = d[3][i]; a[d[3][i]].r = cc[3][i]; } for (i = 0; i < H; i++) { a[dd[3][i]].l = c[3][i]; a[c[3][i]].r = dd[3][i]; } } for (i = 1; i <= n; i++) { int aa = a[i * (m + 2)].id; for (j = 1; j <= m; j++) { aa = a[aa].r; printf("%d", a[aa].val); if (j != m) { printf(" "); } } printf("\n"); } } return 0; }
Codeforces Round #367 (Div. 2)
标签：col 水题 time show 字典 splay nod return isp
原文地址：https://www.cnblogs.com/mountaink/p/10526325.html
