codeforces #602 div2 ABCD

Posted 2020-11-23 mooleetzi

A. Math Problem

Description

 

 

给出n个区间，求一个最短区间使得这个区间与n个区间都有交集

Solution

对$l,r$排序，求出$l_{max}-r_{min}$即可。

做题时被这个卡，真的憨憨。

  1 #include <algorithm>
  2 #include <cctype>
  3 #include <cmath>
  4 #include <cstdio>
  5 #include <cstdlib>
  6 #include <cstring>
  7 #include <iostream>
  8 #include <map>
  9 #include <numeric>
 10 #include <queue>
 11 #include <set>
 12 #include <stack>
 13 #if __cplusplus >= 201103L
 14 #include <unordered_map>
 15 #include <unordered_set>
 16 #endif
 17 #include <vector>
 18 #define lson rt << 1, l, mid
 19 #define rson rt << 1 | 1, mid + 1, r
 20 #define LONG_LONG_MAX 9223372036854775807LL
 21 #define pblank putchar(‘ ‘)
 22 #define ll LL
 23 #define fastIO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
 24 using namespace std;
 25 typedef long long ll;
 26 typedef long double ld;
 27 typedef unsigned long long ull;
 28 typedef pair<int, int> P;
 29 int n, m, k;
 30 const int maxn = 1e5 + 10;
 31 template <class T>
 32 inline T read()
 33 {
 34     int f = 1;
 35     T ret = 0;
 36     char ch = getchar();
 37     while (!isdigit(ch))
 38     {
 39         if (ch == ‘-‘)
 40             f = -1;
 41         ch = getchar();
 42     }
 43     while (isdigit(ch))
 44     {
 45         ret = (ret << 1) + (ret << 3) + ch - ‘0‘;
 46         ch = getchar();
 47     }
 48     ret *= f;
 49     return ret;
 50 }
 51 template <class T>
 52 inline void write(T n)
 53 {
 54     if (n < 0)
 55     {
 56         putchar(‘-‘);
 57         n = -n;
 58     }
 59     if (n >= 10)
 60     {
 61         write(n / 10);
 62     }
 63     putchar(n % 10 + ‘0‘);
 64 }
 65 template <class T>
 66 inline void writeln(const T &n)
 67 {
 68     write(n);
 69     puts("");
 70 }
 71 template <typename T>
 72 void _write(const T &t)
 73 {
 74     write(t);
 75 }
 76 template <typename T, typename... Args>
 77 void _write(const T &t, Args... args)
 78 {
 79     write(t), pblank;
 80     _write(args...);
 81 }
 82 template <typename T, typename... Args>
 83 inline void write_line(const T &t, const Args &... data)
 84 {
 85     _write(t, data...);
 86     puts("");
 87 }
 88 vector<int> l(maxn), r(maxn);
 89 int main(int argc, char const *argv[])
 90 {
 91 #ifndef ONLINE_JUDGE
 92     freopen("in.txt", "r", stdin);
 93     // freopen("out.txt", "w", stdout);
 94 #endif
 95     int t = read<int>();
 96     while (t--)
 97     {
 98         n = read<int>();
 99         for (int i = 0; i < n; i++)
100             l[i] = read<int>(), r[i] = read<int>();
101         if (n == 1)
102         {
103             puts("0");
104             continue;
105         }
106         sort(l.begin(), l.begin() + n);
107         sort(r.begin(), r.begin() + n);
108         writeln(max(l[n - 1] - r[0], 0));
109     }
110     return 0;
111 }

View Code

B. Box

Description

给出一个序列表示当前起始到当前的最大值，问能否构造一个排列。

Solution

模拟即可，并查集维护下一个值。

  1 #include <algorithm>
  2 #include <cctype>
  3 #include <cmath>
  4 #include <cstdio>
  5 #include <cstdlib>
  6 #include <cstring>
  7 #include <iostream>
  8 #include <map>
  9 #include <numeric>
 10 #include <queue>
 11 #include <set>
 12 #include <stack>
 13 #if __cplusplus >= 201103L
 14 #include <unordered_map>
 15 #include <unordered_set>
 16 #endif
 17 #include <vector>
 18 #define lson rt << 1, l, mid
 19 #define rson rt << 1 | 1, mid + 1, r
 20 #define LONG_LONG_MAX 9223372036854775807LL
 21 #define pblank putchar(‘ ‘)
 22 #define ll LL
 23 #define fastIO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
 24 using namespace std;
 25 typedef long long ll;
 26 typedef long double ld;
 27 typedef unsigned long long ull;
 28 typedef pair<int, int> P;
 29 int n, m, k;
 30 const int maxn = 1e5 + 10;
 31 template <class T>
 32 inline T read()
 33 {
 34     int f = 1;
 35     T ret = 0;
 36     char ch = getchar();
 37     while (!isdigit(ch))
 38     {
 39         if (ch == ‘-‘)
 40             f = -1;
 41         ch = getchar();
 42     }
 43     while (isdigit(ch))
 44     {
 45         ret = (ret << 1) + (ret << 3) + ch - ‘0‘;
 46         ch = getchar();
 47     }
 48     ret *= f;
 49     return ret;
 50 }
 51 template <class T>
 52 inline void write(T n)
 53 {
 54     if (n < 0)
 55     {
 56         putchar(‘-‘);
 57         n = -n;
 58     }
 59     if (n >= 10)
 60     {
 61         write(n / 10);
 62     }
 63     putchar(n % 10 + ‘0‘);
 64 }
 65 template <class T>
 66 inline void writeln(const T &n)
 67 {
 68     write(n);
 69     puts("");
 70 }
 71 template <typename T>
 72 void _write(const T &t)
 73 {
 74     write(t);
 75 }
 76 template <typename T, typename... Args>
 77 void _write(const T &t, Args... args)
 78 {
 79     write(t), pblank;
 80     _write(args...);
 81 }
 82 template <typename T, typename... Args>
 83 inline void write_line(const T &t, const Args &... data)
 84 {
 85     _write(t, data...);
 86     puts("");
 87 }
 88 int a[maxn];
 89 int res[maxn];
 90 map<int, int> mp;
 91 set<int> s;
 92 int fa[maxn];
 93 int find(int x)
 94 {
 95     if (fa[x] == 0)
 96         return x;
 97     return fa[x] = find(fa[x]);
 98 }
 99 int main(int argc, char const *argv[])
100 {
101 #ifndef ONLINE_JUDGE
102     freopen("in.txt", "r", stdin);
103     // freopen("out.txt", "w", stdout);
104 #endif
105     int t = read<int>();
106     while (t--)
107     {
108         mp.clear();
109         s.clear();
110         int f = 1;
111         n = read<int>();
112         memset(fa, 0, sizeof(int) * (n + 1));
113         memset(res, 0, sizeof(int) * (n + 1));
114         for (int i = 1; i <= n; i++)
115         {
116             a[i] = read<int>();
117             if (a[i] < i)
118                 f = 0;
119             if (!mp[a[i]])
120             {
121                 res[i] = a[i];
122                 fa[a[i]] = a[i] + 1;
123             }
124             s.emplace(a[i]);
125             mp[a[i]]++;
126         }
127         if (f)
128         {
129             int p = 1;
130             for (int i = 1; i <= n; i++)
131                 if (!res[i])
132                 {
133                     res[i] = find(p);
134                     fa[p] = res[i] + 1;
135                 }
136             for (int i = 1; i <= n; i++)
137                 write(res[i]), pblank;
138             puts("");
139         }
140         else
141             puts("-1");
142     }
143     return 0;
144 }

View Code

 

C. Messy

Description

给出一个括号字符串，每次可以进行操作将一段连续子串翻转。

求一种翻转方式使得翻转后的字符串恰好有k个前缀是匹配字符串。

匹配字符串就是可以正确放入数学表达式的括号串。

Solution

由于没有要求最少交换，直接暴力构造即可。

我选的构造前k-1个前缀是"()",最后一个前缀是这种"(((())))".

  1 #include <algorithm>
  2 #include <cctype>
  3 #include <cmath>
  4 #include <cstdio>
  5 #include <cstdlib>
  6 #include <cstring>
  7 #include <iostream>
  8 #include <map>
  9 #include <numeric>
 10 #include <queue>
 11 #include <set>
 12 #include <stack>
 13 #if __cplusplus >= 201103L
 14 #include <unordered_map>
 15 #include <unordered_set>
 16 #endif
 17 #include <vector>
 18 #define lson rt << 1, l, mid
 19 #define rson rt << 1 | 1, mid + 1, r
 20 #define LONG_LONG_MAX 9223372036854775807LL
 21 #define pblank putchar(‘ ‘)
 22 #define ll LL
 23 #define fastIO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
 24 using namespace std;
 25 typedef long long ll;
 26 typedef long double ld;
 27 typedef unsigned long long ull;
 28 typedef pair<int, int> P;
 29 int n, m, k;
 30 const int maxn = 1e5 + 10;
 31 template <class T>
 32 inline T read()
 33 {
 34     int f = 1;
 35     T ret = 0;
 36     char ch = getchar();
 37     while (!isdigit(ch))
 38     {
 39         if (ch == ‘-‘)
 40             f = -1;
 41         ch = getchar();
 42     }
 43     while (isdigit(ch))
 44     {
 45         ret = (ret << 1) + (ret << 3) + ch - ‘0‘;
 46         ch = getchar();
 47     }
 48     ret *= f;
 49     return ret;
 50 }
 51 template <class T>
 52 inline void write(T n)
 53 {
 54     if (n < 0)
 55     {
 56         putchar(‘-‘);
 57         n = -n;
 58     }
 59     if (n >= 10)
 60     {
 61         write(n / 10);
 62     }
 63     putchar(n % 10 + ‘0‘);
 64 }
 65 template <class T>
 66 inline void writeln(const T &n)
 67 {
 68     write(n);
 69     puts("");
 70 }
 71 template <typename T>
 72 void _write(const T &t)
 73 {
 74     write(t);
 75 }
 76 template <typename T, typename... Args>
 77 void _write(const T &t, Args... args)
 78 {
 79     write(t), pblank;
 80     _write(args...);
 81 }
 82 template <typename T, typename... Args>
 83 inline void write_line(const T &t, const Args &... data)
 84 {
 85     _write(t, data...);
 86     puts("");
 87 }
 88 char s[maxn];
 89 int main(int argc, char const *argv[])
 90 {
 91 #ifndef ONLINE_JUDGE
 92     freopen("in.txt", "r", stdin);
 93     // freopen("out.txt", "w", stdout);
 94 #endif
 95     int t = read<int>();
 96     while (t--)
 97     {
 98         vector<P> res;
 99         res.clear();
100         n = read<int>();
101         k = read<int>();
102         scanf("%s", s);
103         for (int i = 0; i < k - 1; i++)
104         {
105             int curx = i * 2;
106             if (s[curx] == ‘(‘)
107             {
108                 if (s[curx + 1] == ‘(‘)
109                 {
110                     int tox = 0;
111                     for (int j = curx + 2; j < n && !tox; j++)
112                         if (s[j] == ‘)‘)
113                             tox = j;
114                     res.emplace_back(curx + 1, tox);
115                     reverse(s + curx, s + tox + 1);
116                 }
117             }
118             else
119             {
120                 int tox = 0;
121                 for (int j = curx + 1; j < n && !tox; j++)
122                     if (s[j] == ‘(‘)
123                         tox = j;
124                 res.emplace_back(curx, tox);
125                 reverse(s + curx, s + tox + 1);
126                 i--;
127             }
128         }
129         int tox = k - 1 << 1;
130         int left = n - tox >> 1;
131         for (int i = tox, j = 0; i < n; i++, j++)
132         {
133             if (j < left)
134             {
135                 if (s[i] == ‘)‘)
136                 {
137                     int tx = 0;
138                     for (int q = i + 1; q < n && !tx; q++)
139                         if (s[q] == ‘(‘)
140                             tx = q;
141                     res.emplace_back(i, tx);
142                     reverse(s + i, s + tx + 1);
143                 }
144             }
145         }
146         writeln(res.size());
147         for (auto x : res)
148             write_line(x.first + 1, x.second + 1);
149     }
150     return 0;
151 }

View Code

 

D2. Optimal Subsequences (Hard Version)

Description

给出一个长为n的序列。

m次查询，每次查询给出两个值k，p。

要求找出序列长度为k，累加和最大且字典序最小的子序列中的第p个数。

Solution

D1，数据小，暴力开冲。

D2，对原序列按值和idx排序，对查询按找k值排序，离线查询。

对每次查询，先将排序后的值序列前k个的idx插入树状数组，二分找出第p个值得idx，记为当前查询答案。

  1 #include <algorithm>
  2 #include <cctype>
  3 #include <cmath>
  4 #include <cstdio>
  5 #include <cstdlib>
  6 #include <cstring>
  7 #include <iostream>
  8 #include <map>
  9 #include <numeric>
 10 #include <queue>
 11 #include <set>
 12 #include <stack>
 13 #if __cplusplus >= 201103L
 14 #include <unordered_map>
 15 #include <unordered_set>
 16 #endif
 17 #include <vector>
 18 #define lson rt << 1, l, mid
 19 #define rson rt << 1 | 1, mid + 1, r
 20 #define LONG_LONG_MAX 9223372036854775807LL
 21 #define pblank putchar(‘ ‘)
 22 #define ll LL
 23 #define fastIO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
 24 using namespace std;
 25 typedef long long ll;
 26 typedef long double ld;
 27 typedef unsigned long long ull;
 28 typedef pair<int, int> P;
 29 int n, m, k;
 30 const int maxn = 2e5 + 10;
 31 template <class T>
 32 inline T read()
 33 {
 34     int f = 1;
 35     T ret = 0;
 36     char ch = getchar();
 37     while (!isdigit(ch))
 38     {
 39         if (ch == ‘-‘)
 40             f = -1;
 41         ch = getchar();
 42     }
 43     while (isdigit(ch))
 44     {
 45         ret = (ret << 1) + (ret << 3) + ch - ‘0‘;
 46         ch = getchar();
 47     }
 48     ret *= f;
 49     return ret;
 50 }
 51 template <class T>
 52 inline void write(T n)
 53 {
 54     if (n < 0)
 55     {
 56         putchar(‘-‘);
 57         n = -n;
 58     }
 59     if (n >= 10)
 60     {
 61         write(n / 10);
 62     }
 63     putchar(n % 10 + ‘0‘);
 64 }
 65 template <class T>
 66 inline void writeln(const T &n)
 67 {
 68     write(n);
 69     puts("");
 70 }
 71 template <typename T>
 72 void _write(const T &t)
 73 {
 74     write(t);
 75 }
 76 template <typename T, typename... Args>
 77 void _write(const T &t, Args... args)
 78 {
 79     write(t), pblank;
 80     _write(args...);
 81 }
 82 template <typename T, typename... Args>
 83 inline void write_line(const T &t, const Args &... data)
 84 {
 85     _write(t, data...);
 86     puts("");
 87 }
 88 P a[maxn];
 89 struct Node
 90 {
 91     int first, second, idx;
 92     Node(int a, int b, int c) : first(a), second(b), idx(c) {}
 93     Node() {}
 94     bool operator<(const Node &t) const
 95     {
 96         return first < t.first;
 97     }
 98 } q[maxn];
 99 inline int lowbit(int x)
100 {
101     return x & (-x);
102 }
103 struct Ftree
104 {
105     int c[maxn];
106     void add(int x)
107     {
108         while (x < maxn)
109         {
110             ++c[x];
111             x += lowbit(x);
112         }
113     }
114     int query(int x)
115     {
116         int res = 0;
117         while (x)
118         {
119             res += c[x];
120             x -= lowbit(x);
121         }
122         return res;
123     }
124 } tree;
125 int cmp(const P &a, const P &b)
126 {
127     if (a.first == b.first)
128         return a.second < b.second;
129     return a.first > b.first;
130 }
131 int res[maxn], tmp[maxn];
132 inline int judge(int x, int q)
133 {
134     return tree.query(x) >= q;
135 }
136 int main(int argc, char const *argv[])
137 {
138 #ifndef ONLINE_JUDGE
139     freopen("in.txt", "r", stdin);
140     // freopen("out.txt", "w", stdout);
141 #endif
142     n = read<int>();
143     for (int i = 1; i <= n; i++)
144     {
145         int x = read<int>();
146         a[i] = P(x, i);
147     }
148     for (int i = 1; i <= n; i++)
149         tmp[i] = a[i].first;
150     sort(a + 1, a + 1 + n, cmp);
151     m = read<int>();
152     for (int i = 1; i <= m; i++)
153     {
154         int k = read<int>(), p = read<int>();
155         q[i] = Node(k, p, i);
156     }
157     sort(q + 1, q + 1 + m);
158     for (int i = 1, j = 1; i <= m; i++)
159     {
160         while (j <= q[i].first)
161             tree.add(a[j++].second);
162         int l = 1, r = n, cur = 0;
163         while (l <= r)
164         {
165             int mid = l + r >> 1;
166             if (judge(mid, q[i].second))
167                 cur = mid, r = mid - 1;
168             else
169                 l = mid + 1;
170         }
171         res[q[i].idx] = tmp[cur];
172     }
173     for (int i = 1; i <= m; i++)
174         writeln(res[i]);
175     return 0;
176 }

View Code

 

 

five。