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  • AtCoder Beginner Contest 146 ABCD

    A - Can't Wait for Holiday

    Description

    Solution

    模拟水题。

    B - ROT N

    Description

    Solution

    模拟水题。

    C - Buy an Integer

    Description

    Solution

    二分水题,控制一下上界。

      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 ll n, m, k, a, b;
     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 inline ll d(ll x)
     89 {
     90     ll res = 0;
     91     while (x)
     92     {
     93         ++res;
     94         x /= 10;
     95     }
     96     return res;
     97 }
     98 inline int judge(ll x)
     99 {
    100     return a * x + d(x) * b <= n;
    101 }
    102 int main(int argc, char const *argv[])
    103 {
    104 #ifndef ONLINE_JUDGE
    105     // freopen("in.txt", "r", stdin);
    106     // freopen("out.txt", "w", stdout);
    107 #endif
    108 
    109     fastIO;
    110     cin >> a >> b >> n;
    111     ll l = 1, r = 1e9, res = 0;
    112     while (l <= r)
    113     {
    114         ll mid = l + r >> 1;
    115         if (judge(mid))
    116         {
    117             res = mid;
    118             l = mid + 1;
    119         }
    120         else
    121             r = mid - 1;
    122     }
    123     cout << res;
    124     return 0;
    125 }
    View Code

    D - Coloring Edges on Tree

    Description

     给出一个N节点的树,要求一种染色方案使得颜色最少且满足相邻边的颜色不同。

    Solution

    dfs找出儿子最多的一个节点作为根节点。

    从根节点出发开始贪心染色,注意颜色和父节点颜色不同。

      1 #include <algorithm>
      2 #include <cctype>
      3 #include <cmath>
      4 #include <cstdio>
      5 #include <cstdlib>
      6 #include <cstring>
      7 #include <ctime>
      8 #include <iostream>
      9 #include <map>
     10 #include <numeric>
     11 #include <queue>
     12 #include <set>
     13 #include <stack>
     14 #if __cplusplus >= 201103L
     15 #include <unordered_map>
     16 #include <unordered_set>
     17 #endif
     18 #include <vector>
     19 #define lson rt << 1, l, mid
     20 #define rson rt << 1 | 1, mid + 1, r
     21 #define LONG_LONG_MAX 9223372036854775807LL
     22 #define pblank putchar(' ')
     23 #define ll LL
     24 #define fastIO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
     25 using namespace std;
     26 typedef long long ll;
     27 typedef long double ld;
     28 typedef unsigned long long ull;
     29 typedef pair<int, int> P;
     30 int n, m, k;
     31 const int maxn = 1e5 + 10;
     32 template <class T>
     33 inline T read()
     34 {
     35     int f = 1;
     36     T ret = 0;
     37     char ch = getchar();
     38     while (!isdigit(ch))
     39     {
     40         if (ch == '-')
     41             f = -1;
     42         ch = getchar();
     43     }
     44     while (isdigit(ch))
     45     {
     46         ret = (ret << 1) + (ret << 3) + ch - '0';
     47         ch = getchar();
     48     }
     49     ret *= f;
     50     return ret;
     51 }
     52 template <class T>
     53 inline void write(T n)
     54 {
     55     if (n < 0)
     56     {
     57         putchar('-');
     58         n = -n;
     59     }
     60     if (n >= 10)
     61     {
     62         write(n / 10);
     63     }
     64     putchar(n % 10 + '0');
     65 }
     66 template <class T>
     67 inline void writeln(const T &n)
     68 {
     69     write(n);
     70     puts("");
     71 }
     72 template <typename T>
     73 void _write(const T &t)
     74 {
     75     write(t);
     76 }
     77 template <typename T, typename... Args>
     78 void _write(const T &t, Args... args)
     79 {
     80     write(t), pblank;
     81     _write(args...);
     82 }
     83 template <typename T, typename... Args>
     84 inline void write_line(const T &t, const Args &... data)
     85 {
     86     _write(t, data...);
     87     puts("");
     88 }
     89 struct Node
     90 {
     91     int idx, to, nxt;
     92 } edg[maxn << 1];
     93 int head[maxn], son[maxn];
     94 int tot, rt, maxx = 0;
     95 void add(int x, int y, int idx)
     96 {
     97     edg[tot].idx = idx;
     98     edg[tot].to = y;
     99     edg[tot].nxt = head[x];
    100     head[x] = tot++;
    101 }
    102 int col[maxn];
    103 void dfs(int u, int pre)
    104 {
    105     for (int i = head[u]; ~i; i = edg[i].nxt)
    106     {
    107         int v = edg[i].to;
    108         if (v == pre)
    109             continue;
    110         dfs(v, u);
    111         ++son[u];
    112     }
    113     if (son[u] + (u != 1) > maxx)
    114     {
    115         maxx = son[u] + (u != 1);
    116         rt = u;
    117     }
    118 }
    119 void dfs2(int u, int fa, int pre)
    120 {
    121     int x = 1;
    122     if (x == pre)
    123         ++x;
    124     if (x > maxx)
    125         x = 1;
    126     for (int i = head[u]; ~i; i = edg[i].nxt)
    127     {
    128         int v = edg[i].to;
    129         int num = edg[i].idx;
    130         if (col[num] || v == fa)
    131             continue;
    132         col[num] = x;
    133         ++x;
    134         if (x > maxx)
    135             x = 1;
    136         if (x == pre)
    137             ++x;
    138         if (x > maxx)
    139             x = 1;
    140     }
    141     for (int i = head[u]; ~i; i = edg[i].nxt)
    142     {
    143         int v = edg[i].to;
    144         int num = edg[i].idx;
    145         if (v != fa)
    146             dfs2(v, u, col[num]);
    147     }
    148 }
    149 int main(int argc, char const *argv[])
    150 {
    151 #ifndef ONLINE_JUDGE
    152     // freopen("in.txt", "r", stdin);
    153     // freopen("out.txt", "w", stdout);
    154 #endif
    155     srand(time(nullptr));
    156     n = read<int>();
    157     memset(head, -1, sizeof(int) * (n + 1));
    158     for (int i = 0; i < n - 1; i++)
    159     {
    160         int x = read<int>(), y = read<int>();
    161         add(x, y, i);
    162         add(y, x, i);
    163     }
    164     dfs(1, 0);
    165     writeln(maxx);
    166     dfs2(rt, 0, 0);
    167     for (int i = 0; i < n - 1; i++)
    168         writeln(col[i]);
    169     return 0;
    170 }
    View Code
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  • 原文地址:https://www.cnblogs.com/mooleetzi/p/11954563.html
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