2018-2019 ACM-ICPC, Asia Seoul Regional Contest

ProblemA Circuits

Solved.

题意：

有$n$个矩形，可以放两条平行与$x$轴的线，求怎么放置两条无线长的平行于$x$轴的线，使得他们与矩形相交个数最多

如果一个矩形同时与两条线相交，只算一次。

思路：

离散化后枚举一根线，另一根线用线段树维护，扫描线思想

 #include <bits/stdc++.h>
 using namespace std;

 #define N 400010
 int n;
 int b[N];
 int x[N], y[N];
 vector <int> in[N], out[N];
 int ans[N];

 namespace SEG
 {
     struct node
     {
         int Max, lazy;
         node () {}
         node (int Max, int lazy) : Max(Max), lazy(lazy) {}
         void init() { Max = lazy = ; }
         void add(int x)
         {
             Max += x;
             lazy += x;
         }
         node operator + (const node &other) const
         {
             node res; res.init();
             res.Max = max(Max, other.Max);
             return res;
         }
     }a[N << ];
     void build(int id, int l, int r)
     {
         a[id].init();
         if (l == r)
             return;
         int mid = (l + r) >> ;
         build(id << , l, mid);
         build(id <<  | , mid + , r);
     }
     void pushdown(int id)
     {
         if (!a[id].lazy) return;
         a[id << ].add(a[id].lazy);
         a[id <<  | ].add(a[id].lazy);
         a[id].lazy = ;
     }
     void update(int id, int l, int r, int ql, int qr, int x)
     {
         if (l >= ql && r <= qr)
         {
             a[id].add(x);
             return;
         }
         int mid = (l + r) >> ;
         pushdown(id);
         if (ql <= mid) update(id << , l, mid, ql, qr, x);
         if (qr > mid) update(id <<  | , mid + , r, ql, qr, x);
         a[id] = a[id << ] + a[id <<  | ];
     }
     int query(int id, int l, int r, int pos)
     {
         if (l == r) return a[id].Max;
         int mid = (l + r) >> ;
         pushdown(id);
         if (pos <= mid) return query(id << , l, mid, pos);
         else return query(id <<  | , mid + , r, pos);
     }
 }

 void Hash()
 {
     sort(b + , b +  + b[]);
     b[] = unique(b + , b +  + b[]) - b - ;
     for (int i = ; i <= n; ++i) x[i] = lower_bound(b + , b +  + b[], x[i]) - b;
     for (int i = ; i <= n; ++i) y[i] = lower_bound(b + , b +  + b[], y[i]) - b;
 }

 int main()
 {
     while (scanf("%d", &n) != EOF)
     {
         b[] = ;
         for (int i = ; i < N; ++i)
             in[i].clear(), out[i].clear();
         for (int i = , tmp; i <= n; ++i)
         {
             scanf("%d%d%d%d", &tmp, y + i, &tmp, x + i);
 //            cout << x[i] << " " << y[i] << endl;
             b[++b[]] = x[i];
             b[++b[]] = y[i];
         }
         Hash();
         SEG::build(, , b[]);
         for (int i = ; i <= n; ++i)
         {
             in[x[i]].push_back(i);
             out[y[i]].push_back(i);
             SEG::update(, , b[], x[i], y[i], );
         }
         int res = ;
         for (int i = ; i <= b[]; ++i)
             ans[i] = SEG::query(, , b[], i);
         for (int i = ; i <= b[]; ++i)
         {
             for (auto it : in[i])
                 SEG::update(, , b[], x[it], y[it], -);
             res = max(res, ans[i] + SEG::a[].Max);
             for (auto it : out[i])
                 SEG::update(, , b[], x[it], y[it], );
         }
         printf("%d\n", res);
     }
     return ;
 }

Problem D Go Latin

Solved.

按题意模拟。

 #include <bits/stdc++.h>
 using namespace std;

 int t;
 map <string, string> mp;

 int main()
 {
     mp["a"] = "as";
     mp["i"] = "ios";
     mp["y"] = "ios";
     mp["l"] = "les";
     mp["n"] = "anes";
     mp["ne"] = "anes";
     mp["o"] = "os";
     mp["r"] = "res";
     mp["t"] = "tas";
     mp["u"] = "us";
     mp["v"] = "ves";
     mp["w"] = "was";
     ios::sync_with_stdio(false);
     cin.tie(); cout.tie();
     cin >> t;
     string s;
     while (t--)
     {
         cin >> s;
         int len = s.size();
         string tmp = "";
         tmp += s[len - ];
         if (mp.find(tmp) != mp.end())
         {
             s.erase(s.begin() + len - );
             s += mp[tmp];
         }
         else
         {
             tmp = "";
             tmp += s[len - ];
             tmp += s[len - ];
             if (mp.find(tmp) != mp.end())
             {
                 s.erase(s.begin() + len - , s.end());
                 s += mp[tmp];
             }
             else
                 s += "us";
         }
         cout << s << "\n";
     }
     return ;
 }

ProblemE LED

Upsolved.

题意：求一个分段函数的最小最大值

思路：

注意$v = 0$的时候，其余部分二分，贪心验证

 #include <bits/stdc++.h>
 using namespace std;

 #define ll long long
 #define N 300010
 struct node
 {
     ll v, l;
     void scan()
     {
         scanf("%lld%lld", &v, &l);
         l *= 10ll;
     }
     bool operator < (const node &other) const { return v < other.v; }
 }a[N];
 int n;

 bool check(ll x)
 {
     int i;
     ll l1, l2;
     for (i = ; i <= n; ++i)
     {
         if (a[i].l > x)
             break;
     }
     if (i > n) return true;
     ll Max = , Min = (ll)1e11;
     for (; i <= n; ++i)
     {
         Max = max(Max, a[i].l);
         Min = min(Min, a[i].l);
         if (Max - (Max + Min) /  > x)
             break;
         l1 = (Max + Min) / ;
     }
     if (i > n) return true;
     Max = , Min = (ll)1e11;
     for (; i <= n; ++i)
     {
         Max = max(Max, a[i].l);
         Min = min(Min, a[i].l);
         if (Max - (Max + Min) /  > x)
             break;
         l2 = (Max + Min) / ;
     }
     //cout << x << " " << l1 << " " << l2 << endl;
     if (i <= n) return false;
     return l2 >= l1;
 }

 int main()
 {
     while (scanf("%d", &n) != EOF)
     {
         for (int i = ; i <= n; ++i) a[i].scan();
         sort(a + , a +  + n);
         ll base = ;
         int j = ;
         for (int i = ; i <= n; ++i)
         {
             if (a[i].v == )
                 base = max(base, a[i].l);
             else
                 a[++j] = a[i];
         }
         n = j;
         ll l = , r = (ll)1e11, res = -;
         while (r - l >= )
         {
             ll mid = (l + r) >> ;
             if (check(mid))
             {
                 res = mid;
                 r = mid - ;
             }
             else
                 l = mid + ;
         }
         res = max(res, base);
         ll x = res % ;
         res /= ;
         printf("%lld.%lld\n", res, x);
     }
     return ;
 }

ProblemF Parentheses

Upsolved.

题意：给出一个表达式，先判合法性，再判是否有括号冗余或者缺少。

思路：

直接判吧，数据里面有一个空行，不要多组读入。。。

vp的时候以为有(-b)这种东西，实际上没有。

 #include <bits/stdc++.h>
 using namespace std;

 #define N 1000100
 char s[N];
 int id[N], len;

 bool ok()
 {
     len = strlen(s + );
     for (int i = ; i <= len; ++i)
     {
         if (isalpha(s[i]))
             id[i] = ;
         else if (s[i] == '+' ||
                  s[i] == '-' ||
                  s[i] == '*' ||
                  s[i] == '/' ||
                  s[i] == '%')
                  id[i] = ;
         else if (s[i] == '(')
             id[i] = ;
         else
             id[i] = ;
     }
     if (id[] ==  || id[len] == ) return false;
     if (id[] ==  || id[len] == ) return false;
     if (len ==  && id[] == )
         return true;
     for (int i = , x, y; i <= len; ++i)
     {
         x = id[i - ], y = id[i];
         if (x ==  && y == )
             return false;
         if (x ==  && y == )
             return false;
         if (x ==  && y == )
             return false;
         if (x ==  && y == )
             return false;
         if (x ==  && y == )
             return false;
         if (x ==  && y == )
             return false;
         if (x ==  && y == )
             return false;
         if (x ==  && y == )
             return false;
     }
     int cnt = ;
     for (int i = ; i <= len; ++i)
     {
         if (id[i] == )
             ++cnt;
         else if (id[i] == )
         {
             if (cnt == )
                 return false;
             --cnt;
         }
     }
     return cnt == ;
 }

     // 0 alpha
     // 1 + - * / %
     // 2 (
     // 3 )

 bool work()
 {
     if (len == ) return true;
     int cnt = , sym = ;
     for (int i = ; i <= len; ++i)
         sym += (id[i] == );
     stack <int> sta;
     for (int i = ; i <= len; ++i)
     {
         if (id[i] == )
             continue;
         else if (id[i] <= )
             sta.push(id[i]);
         else if (id[i] == )
         {
             int flag = ;
             while (!sta.empty())
             {
                 int top = sta.top(); sta.pop();
                 if (top == )
                     break;
                 if (top == )
                     ++flag;
             }
             if (flag == ) ++cnt;
             else
                 return false;
         }
     }
     return cnt == sym - ;
 }

 int main()
 {
     gets(s + );
     {
         len = strlen(s + );
         //if (len == 0) continue;
         int now = ;
         for (int i = ; i <= len; ++i)
             if (s[i] != ' ')
                 s[++now] = s[i];
         s[++now] = ;
 //        cout << s + 1 << endl;
         if (!ok()) puts("error");
         else
             puts(work() ? "proper" : "improper");
     }
     return ;
 }

ProblemK TV Show Game

Solved.

题意：$k$个灯， $n$个人， 每个人猜三个位置的灯的颜色， 要求构造出使得所有人都猜对两个或两个以上数量的灯的颜色的序列。

思路：枚举每个灯的状态， 2-SAT求解。

(2-SAT 不熟练啊 果然还是需要练一练)

 #include<bits/stdc++.h>

 using namespace std;

 const int maxn = 1e4 + ;

 struct node{
     int lamp1, lamp2, lamp3;
     int color1, color2, color3;
 }arr[maxn];

 int k, n;
 int vis[maxn << ];
 vector<int>People[maxn], lamp[maxn];//R 0 B 1
 int Stack[maxn], top;

 void Init()
 {
     for(int i = ; i < maxn; ++i) People[i].clear();
     for(int i = ; i < maxn; ++i) lamp[i].clear();
 }

 bool DFS(int u)
 {
     if(vis[u ^ ]) return false;
     if(vis[u]) return true;
     vis[u] = ;
     Stack[top++] = u;
     for(auto it : lamp[u])
     {
         for(auto v : People[it]) if(v != u)
         {
             if(!DFS(v ^ )) return false;
         }
     }
     return true;
 }

 bool Twosat(int n)
 {
     memset(vis, , sizeof vis);
     for(int i = ; i <= n; i += )
     {
         if(vis[i] || vis[i ^ ]) continue;
         top = ;
         if(!DFS(i))
         {
             while(top) vis[Stack[--top]] = ;
             if(!DFS(i ^ )) return false;
         }
     }
     return true;
 }

 int main()
 {
     while(~scanf("%d %d", &k, &n))
     {
         Init();
         for(int i = ; i <= n; ++i)
         {
             char a, b, c;
             scanf("%d %c %d %c %d %c", &arr[i].lamp1, &a, &arr[i].lamp2, &b, &arr[i].lamp3, &c);

             arr[i].color1 = (a == 'B');
             arr[i].color2 = (b == 'B');
             arr[i].color3 = (c == 'B');

             People[i].push_back(arr[i].lamp1 *  + (arr[i].color1 ^ ));
             lamp[arr[i].lamp1 *  + (arr[i].color1 ^ )].push_back(i);

             People[i].push_back(arr[i].lamp2 *  + (arr[i].color2 ^ ));
             lamp[arr[i].lamp2 *  + (arr[i].color2 ^ )].push_back(i);

             People[i].push_back(arr[i].lamp3 *  + (arr[i].color3 ^ ));
             lamp[arr[i].lamp3 *  + (arr[i].color3 ^ )].push_back(i);
         }
         if(Twosat( * k))
         {
             for(int i = ; i <= k; ++i)
             {
                 if(vis[ * i]) putchar('R');
                 else putchar('B');
             }
             putchar('\n');
         }
         else puts("-1");
     }
     return ;
 }

ProblemL Working Plan

Solved.

题意：$m$个人， $n$天， 每个人一旦开始工作需要连续工作$w$天， 工作完w天后需要休息$h$天， 每天需要$d_i$个人， 每个人要求工作$W_i$天， 求方案。

思路：贪心， 优先使得剩余工作天数多的人工作。

 #include<bits/stdc++.h>

 using namespace std;

 typedef long long ll;

 const int maxn = 2e3 + ;

 struct node{
     int id;
     int st, ed;
     int workday;
     node(){}
     node(int id, int st, int ed, int workday): id(id), st(st), ed(ed), workday(workday){}
     bool operator < (const node  &other) const{
         if(st != other.st) return st > other.st;
         else return workday < other.workday;
     }
 };

 bool cmp(node a, node b)
 {
     return a.workday < b.workday;
 }

 int n, m,w, h;
 int W[maxn], d[maxn];
 priority_queue<node>q;
 priority_queue<int, vector<int>, greater<int> >Day;
 vector<int>ans[maxn];
 node Stack[maxn];
 int top;

 void Init()
 {
     while(!q.empty()) q.pop();
     while(!Day.empty()) Day.pop();
     for(int i = ; i <= m; ++i) ans[i].clear();
 }

 int main()
 {
     while(~scanf("%d %d %d %d", &m, &n, &w, &h))
     {
         Init();
         for(int i = ; i <= m; ++i) scanf("%d", W + i);
         for(int i = ; i <= n; ++i) scanf("%d", d + i);
         for(int i = ; i <= m; ++i) q.push(node(i, , , W[i]));
         bool flag = true;
         int cnt = ;
         top = ;
         for(int i = ; i <= n; ++i)
         {
             while(!Day.empty())
             {
                 if(Day.top() < i) Day.pop();
                 else break;
             }
             int daysize = Day.size();
             if(daysize > d[i])
             {
                 flag = false;
                 break;
             }
             if(daysize == d[i]) continue;
             while(!q.empty())
             {
                 if(q.top().st <= i)
                 {
                     Stack[++top] = q.top();
                     q.pop();
                 }
                 else
                 {
                     break;
                 }
             }
             sort(Stack + , Stack +  + top, cmp);
             while(Day.size() < d[i])
             {
                 if(top < )
                 {
                     flag = false;
                     break;
                 }
                 node tmp = Stack[top--];
                 tmp.st = i;
                 ans[tmp.id].push_back(i);
                 tmp.st += w;
                 Day.push(tmp.st - );
                 tmp.st += h;
                 tmp.workday -= min(w, n - i + );
                 if(tmp.workday) q.push(tmp);
                 else cnt++;
             }
             if(Day.size() != d[i]) flag= false;
             if(flag == false) break;
         }
         if(cnt != m) flag = false;
         if(flag)
         {
             puts("");
             for(int i = ; i <= m; ++i) for(int j = , len = ans[i].size(); j < len; ++j) printf("%d%c", ans[i][j], " \n"[j == len - ]);
         }
         else puts("-1");
     }
     return ;
 }