• “玲珑杯”ACM比赛 Round #12题解&源码

    我能说我比较傻么！就只能做一道签到题，没办法，我就先写下A题的题解&源码吧，把官方给出的题解贴出来！

                                        A -- Niro plays Galaxy Note 7                   

                    Time Limit：1s 

                           Memory Limit：128MByte                   

    DESCRIPTION
           
               

    Niro, a lovely girl, has bought a Galaxy Note 7 and wants to destroy cities. There are N cities numbered 1... N on a line and each pair of adjacent cities has distance 1. Galaxy Note 7 has its explosion radius R. Niro puts her Galaxy Note 7 in city X and city i will be destroyed if (|X−i|≤R)

    .You must tell Niro how many cities wil be destroyed.

                       
       

       

           
    INPUT
           
               
    The first line contains a positive integer T, the number of test cases.
    Each of the following T lines contains three integers N, R, X    .
    OUTPUT
                
    Tlines.Each line contains one integer, the answer.
          
    SAMPLE INPUT
               
    3
    100 5 23
    100 8 36
    100 9 99
    SAMPLE OUTPUT
                
    11
    17
    11
        
    HINT
               
    1≤T,N≤100

    0≤R≤100
    1≤X≤N                 

           
    SOLUTION
    “玲珑杯”ACM比赛 Round #12 
    题目链接：http://www.ifrog.cc/acm/problem/1106?contest=1014&no=0
    分析：这道题就是所谓的签到题，不是很难，能够摧毁的城市是区间 [max(1,X−i),min(X+i,N)]，直接输出min(X+i,N)−max(1,X−i)+1即可，题解的那种方式看不太懂，可能是因为我自己没学C++STL，其实就是以一个点为中心，向左区间和右区间分别延伸R个单位，如果超过N或小于0终止！
    下面给出AC代码：
     1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int main()
 4 {
 5     int T;
 6     int n,r,x;
 7     int a[1010];
 8     while(scanf("%d",&T)!=EOF)
 9     {
10         while(T--)
11         {
12             scanf("%d%d%d",&n,&r,&x);
13             memset(a,0,sizeof(a));
14             int ans=0;
15             if(x<=n)
16             {
17                 for(int i=x;i<=x+r;i++)
18                 {
19                     if(i<=n)
20                     {
21                         a[i]=1;
22                     }
23                 }
24                 for(int i=x;i>=x-r;i--)
25                 {
26                     if(i>=0)
27                     {
28                         a[i]=1;
29                     }
30                 }
31                 for(int i=1;i<=n;i++)
32                     if(a[i])
33                     ans++;
34                     printf("%d
",ans);
35             }
36         }
37     }
38     return 0;
39 }

     给出官方的STL解法：

     1 #include <cstdio>
 2 #include <algorithm>
 3 int T, N, R, X;
 4 int main()
 5 {
 6     for (scanf("%d", &T); T--; )
 7     {
 8         scanf("%d%d%d", &N, &R, &X);
 9         printf("%d
", std::min(N, X + R) - std::max(1, X - R) + 1);
10     }
11     return 0;
12 }

    题目链接：http://www.ifrog.cc/acm/problem/1107?contest=1014&no=1

    题解：

     下面给出AC代码：

      1 #include <cstdio>
  2 #include <queue>
  3 #include <vector>
  4 #include <algorithm>
  5 const int INF = 1000000000;
  6 class Heap
  7 {
  8     private :
  9         std::priority_queue < int, std::vector < int >, std::greater < int > > inc, dec;
 10         void BaseClear()
 11         {
 12             while (!dec.empty() && inc.top() == dec.top())
 13             {
 14                 inc.pop();
 15                 dec.pop();
 16             }
 17         }
 18     public :
 19         int top()
 20         {
 21             BaseClear();
 22             return inc.top();
 23         }
 24         void del(int x)
 25         {
 26             dec.push(x);
 27         }
 28         void push(int x)
 29         {
 30             inc.push(x);
 31         }
 32         void clear()
 33         {
 34             while (!inc.empty())
 35                 inc.pop();
 36             while (!dec.empty())
 37                 dec.pop();
 38         }
 39         bool empty()
 40         {
 41             BaseClear();
 42             return inc.empty();
 43         }
 44 }
 45 Q0, Q1;
 46 int TC, f0[200001], f1[200001], *F0 = f0 + 100000, *F1 = f1 + 100000, N, C0, C1, N0, N1, E0, E1, TAG0, TAG1;
 47 void forward(char option)
 48 {
 49     if (option == '0')
 50     {
 51         F0--;
 52         F0[1] = (Q1.empty() ? INF : Q1.top()) + TAG1 - TAG0;
 53         E0++;
 54         Q0.push(F0[1]);
 55         while (E0 >= N0)
 56             Q0.del(F0[E0--]);
 57         E1 = 0;
 58         Q1.clear();
 59     }
 60     else if (option == '1')
 61     {
 62         F1--;
 63         F1[1] = (Q0.empty() ? INF : Q0.top()) + TAG0 - TAG1;
 64         E1++;
 65         Q1.push(F1[1]);
 66         while (E1 >= N1)
 67             Q1.del(F1[E1--]);
 68         E0 = 0;
 69         Q0.clear();
 70     }
 71     else
 72     {
 73         F0--;
 74         F0[1] = (Q1.empty() ? INF : Q1.top()) + TAG1 - TAG0;
 75         E0++;
 76         F1--;
 77         F1[1] = (Q0.empty() ? INF : Q0.top()) + TAG0 - TAG1;
 78         E1++;
 79         Q0.push(F0[1]);
 80         Q1.push(F1[1]);
 81         while (E0 >= N0)
 82             Q0.del(F0[E0--]);
 83         while (E1 >= N1)
 84             Q1.del(F1[E1--]);
 85         TAG0 += C0;
 86         TAG1 += C1;
 87     }
 88 }
 89 int main()
 90 {
 91     for (scanf("%d", &TC); TC--; )
 92     {
 93         F0 = f0 + 100000;
 94         F1 = f1 + 100000;
 95         TAG0 = TAG1 = 0;
 96         Q0.clear();
 97         Q1.clear();
 98         E0 = E1 = 0;
 99         scanf("%d%d%d%d%d", &N, &C0, &C1, &N0, &N1);
100         char c = getchar();
101         while (c != '0' && c != '1' && c != '?')
102             c = getchar();
103         if (c == '0')
104         {
105             F0[E0 = 1] = 0;
106             Q0.push(0);
107         }
108         else if (c == '1')
109         {
110             F1[E1 = 1] = 0;
111             Q1.push(0);
112         }
113         else
114         {
115             F0[E0 = 1] = C0;
116             F1[E1 = 1] = C1;
117             Q0.push(C0);
118             Q1.push(C1);
119         }
120         for (int i = 1; i < N; i++)
121             forward(getchar());
122         int ans = 1000000001;
123         if (!Q0.empty())
124             ans = std::min(ans, Q0.top() + TAG0);
125         if (!Q1.empty())
126             ans = std::min(ans, Q1.top() + TAG1);
127         printf("%d
", ans);
128     }
129     return 0;
130 }

    题目链接：http://www.ifrog.cc/acm/problem/1108?contest=1014&no=2

    题解：

    下面给出AC代码：

     1 #include <bits/stdc++.h>
 2 const int MOD = 1234321237;
 3 int F[100001], N, G, a[1000], w[1000];
 4 int gcd(int x, int y)
 5 {
 6     int r;
 7     while (y)
 8     {
 9         r = x % y;
10         x = y;
11         y = r;
12     }
13     return x;
14 }
15 void DP(int x, int y)
16 {
17     std::vector < int > Div;
18     for (int i = 1; i * i <= x; i++)
19         if (x % i == 0)
20         {
21             Div.push_back(i);
22             if (i * i < x)
23                 Div.push_back(x / i);
24         }
25     std::sort(Div.begin(), Div.end());
26     int L = Div.size();
27     std::vector < int > Use(L, 0);
28     for (int i = L - 1; ~i; i--)
29     {
30         Use[i] = y / Div[i];
31         for (int j = i + 1; j < L; j++)
32             if (Div[j] % Div[i] == 0)
33                 Use[i] -= Use[j];
34     }
35     for (int i = G; ~i; i--)
36     {
37         F[i] = 0;
38         for (int j = 0; j < L && Div[j] <= i; j++)
39             F[i] = (F[i] + (long long)F[i - Div[j]] * Use[j]) % MOD;
40     }
41 }
42 int main()
43 {
44     scanf("%d%d", &N, &G);
45     for (int i = 0; i < N; i++)
46         scanf("%d", a + i);
47     for (int i = 0; i < N; i++)
48         scanf("%d", w + i);
49     F[0] = 1;
50     for (int i = 0; i < N; i++)
51         DP(a[i], w[i]);
52     printf("%d
", F[G]);
53     return 0;
54 }

    题目链接：http://www.ifrog.cc/acm/problem/1109?contest=1014&no=3

    题解：

    下面给出AC代码：

     1 #include <cstdio>
 2 const long long MOD = 1234321237;
 3 long long POWER(long long a, long long b)
 4 {
 5     long long r = 1;
 6     for (; b; b >>= 1)
 7     {
 8         if (b & 1)
 9             r = r * a % MOD;
10         a = a * a % MOD;
11     }
12     return r;
13 }
14 long long N;
15 int T;
16 int main()
17 {
18     for (scanf("%d", &T); T--; )
19     {
20         scanf("%lld", &N);
21         long long F = POWER(4, N - 1) * 3 - POWER(3, N - 1) * 2;
22         long long G = POWER(4, N - 1) * (((N % MOD * 9) - 69) % MOD) + POWER(3, N - 1) * (((N % MOD * 8) + 52) % MOD);
23         G %= MOD;
24         F %= MOD;
25         G %= MOD;
26         F += MOD;
27         G += MOD;
28         F %= MOD;
29         G %= MOD;
30         if (G & 1)
31             G += MOD;
32         G >>= 1;
33         printf("%lld %lld
", F, G);
34     }
35     return 0;
36 }

    题目链接：http://www.ifrog.cc/acm/problem/1110?contest=1014&no=4

    题解：

    下面给出AC代码：

      1 #include <cstdio>
  2 #include <vector>
  3 #include <algorithm>
  4 std::vector < int > E[100001], col[100001];
  5 std::vector < std::pair < int, int > > inc[100002], dec[100002];
  6 int N, q[100001], left[100001], right[100001], size[100001], BeiZeng[17][100001], *fa = BeiZeng[0], LOG; // left : DFN; right maximum DFN in its subtree
  7 std::vector < int >::iterator ue[100001];
  8 void DFS()
  9 {
 10     int D = 1, TIME = 1;
 11     q[1] = 1;
 12     ue[1] = E[1].begin();
 13     left[1] = right[1] = 1;
 14     while (D)
 15     {
 16         if (ue[D] != E[q[D]].end() && *ue[D] == fa[q[D]])
 17             ue[D]++;
 18         if (ue[D] != E[q[D]].end())
 19         {
 20             int To = *ue[D]++;
 21             fa[To] = q[D];
 22             left[To] = right[To] = ++TIME;
 23             q[++D] = To;
 24             ue[D] = E[To].begin();
 25         }
 26         else
 27         {
 28             if (D > 1)
 29                 right[q[D - 1]] = right[q[D]];
 30             D--;
 31         }
 32     }
 33     for (int i = 1; i <= N; i++)
 34         size[i] = right[i] - left[i] + 1;
 35     while (2 << LOG < N)
 36         LOG++;
 37     for (int i = 1; i <= LOG; i++)
 38         for (int j = 1; j <= N; j++)
 39             BeiZeng[i][j] = BeiZeng[i - 1][BeiZeng[i - 1][j]];
 40 }
 41 int lowest(int u, int v)
 42 {
 43     for (int i = LOG; ~i; i--)
 44         if (BeiZeng[i][u] && size[BeiZeng[i][u]] < size[v])
 45             u = BeiZeng[i][u];
 46     return u;
 47 }
 48 inline void bar(int u, int d, int l, int r)
 49 {
 50     inc[u].push_back(std::make_pair(l, r));
 51     if (d < N)
 52         dec[d + 1].push_back(std::make_pair(l, r));
 53 }
 54 void conflict(int u, int v)
 55 {
 56     if (size[u] < size[v])
 57         std::swap(u, v);
 58     if (left[u] <= left[v] && right[v] <= right[u]) // u is v's ancestor
 59     {
 60         int lw = lowest(v, u);
 61         if (left[lw] > 1)
 62         {
 63             bar(left[v], right[v], 1, left[lw] - 1);
 64             bar(1, left[lw] - 1, left[v], right[v]);
 65         }
 66         if (right[lw] < N)
 67         {
 68             bar(left[v], right[v], right[lw] + 1, N);
 69             bar(right[lw] + 1, N, left[v], right[v]);
 70         }
 71     }
 72     else
 73     {
 74         bar(left[u], right[u], left[v], right[v]);
 75         bar(left[v], right[v], left[u], right[u]);
 76     }
 77 }
 78 int MIN[262145], TAG[262145], NUM[262145]; // NUM[] : the number of elements which reach MIN[]
 79 void INC(int p, int l, int r, int L, int R, int w)
 80 {
 81     if (L <= l && r <= R)
 82     {
 83         MIN[p] += w;
 84         TAG[p] += w;
 85         return;
 86     }
 87     if (TAG[p])
 88     {
 89         MIN[p + p] += TAG[p];
 90         MIN[p + p + 1] += TAG[p];
 91         TAG[p + p] += TAG[p];
 92         TAG[p + p + 1] += TAG[p];
 93         TAG[p] = 0;
 94     }
 95     int m = (l + r) >> 1;
 96     if (L <= m)
 97         INC(p + p, l, m, L, R, w);
 98     if (R > m)
 99         INC(p + p + 1, m + 1, r, L, R, w);
100     MIN[p] = std::min(MIN[p + p], MIN[p + p + 1]);
101     NUM[p] = (MIN[p + p] == MIN[p] ? NUM[p + p] : 0) + (MIN[p + p + 1] == MIN[p] ? NUM[p + p + 1] : 0);
102 }
103 inline int ZERONUM()
104 {
105     return MIN[1] == 0 ? NUM[1] : 0;
106 }
107 long long ANS;
108 void Treeinit(int p = 1, int l = 1, int r = N)
109 {
110     NUM[p] = r - l + 1;
111     if (l < r)
112     {
113         int m = (l + r) >> 1;
114         Treeinit(p + p, l, m);
115         Treeinit(p + p + 1, m + 1, r);
116     }
117 }
118 int main()
119 {
120     scanf("%d", &N);
121     for (int i = 1, u, v; i < N; i++)
122     {
123         scanf("%d%d", &u, &v);
124         E[u].push_back(v);
125         E[v].push_back(u);
126     }
127     for (int i = 1, c; i <= N; i++)
128     {
129         scanf("%d", &c);
130         col[c].push_back(i);
131     }
132     DFS();
133     for (int i = 1; i <= N; i++)
134         for (std::vector < int >::iterator x = col[i].begin(); x != col[i].end(); x++)
135             for (std::vector < int >::iterator y = x + 1; y != col[i].end(); y++)
136                 conflict(*x, *y);
137     Treeinit();
138     for (int i = 1; i <= N; i++)
139     {
140         for (std::vector < std::pair < int, int > >::iterator j = inc[i].begin(); j != inc[i].end(); j++)
141             INC(1, 1, N, j -> first, j -> second, 1);
142         for (std::vector < std::pair < int, int > >::iterator j = dec[i].begin(); j != dec[i].end(); j++)
143             INC(1, 1, N, j -> first, j -> second, -1);
144         ANS += ZERONUM();
145     }
146     printf("%lld
", (ANS - N) >> 1);
147     return 0;
148 }
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  • 原文地址：https://www.cnblogs.com/ECJTUACM-873284962/p/6574397.html
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