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  • 2017-2018 ACM-ICPC German Collegiate Programming Contest (GCPC 2017) Solution

    A. Drawing Borders

    Unsolved.

    B. Buildings

    Unsolved.

    C. Joyride

    Upsolved.

    题意:

    在游乐园中,有n个游玩设施,有些设施之间有道路,给出每个设施需要花费的时间和价格,以及经过每条道路的时间,求从设施1出发恰好回到设施1恰好花费时间x的最小花费。

    思路:

    $dist[i][j] 表示 第i个时刻 到第j个点的最小花费  从<t[1], 1> -> <x, 1>$

    跑最短路即可

     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 
     4 #define N 3010
     5 #define ll long long
     6 #define INF 0x3f3f3f3f3f3f3f3f
     7 int X, n, m, T, t[N], p[N]; 
     8 vector <int> G[N];
     9 
    10 struct node
    11 {
    12     int x, to; ll w;
    13     node () {}
    14     node (int x, int to, ll w) : x(x), to(to), w(w) {}
    15     bool operator < (const node &r) const { return w > r.w; }
    16 };
    17 
    18 bool used[N][N]; 
    19 ll dist[N][N];
    20 void Dijkstra()
    21 {
    22     for (int i = 1; i <= X; ++i) for (int j = 1; j <= n; ++j) dist[i][j] = INF, used[i][j] = 0;
    23     priority_queue <node> q; q.emplace(t[1], 1, 0); dist[t[1]][1] = p[1]; 
    24     while (!q.empty())
    25     {
    26         int x = q.top().x, u = q.top().to; q.pop();
    27         if (x > X) continue;
    28         if (used[x][u]) continue;
    29         used[x][u] = 1;
    30         if (!used[x + t[u]][u] && dist[x + t[u]][u] > dist[x][u] + p[u]) 
    31         {
    32             dist[x + t[u]][u] = dist[x][u] + p[u];
    33             q.emplace(x + t[u], u, dist[x + t[u]][u]);
    34         }
    35         for (auto v : G[u]) if (!used[x + t[v] + T][v] && dist[x + t[v] + T][v] > dist[x][u] + p[v])
    36         {
    37             dist[x + t[v] + T][v] = dist[x][u] + p[v];
    38             q.emplace(x + t[v] + T, v, dist[x + t[v] + T][v]);
    39         }
    40     }    
    41 }
    42 
    43 int main()
    44 {
    45     while (scanf("%d", &X) != EOF)
    46     {
    47         scanf("%d%d%d", &n, &m, &T);
    48         for (int i = 1, u, v; i <= m; ++i)
    49         {
    50             scanf("%d%d", &u, &v);
    51             G[u].push_back(v);
    52             G[v].push_back(u);
    53         }
    54         for (int i = 1; i <= n; ++i) scanf("%d%d", t + i, p + i);
    55         Dijkstra();
    56         if (dist[X][1] == INF) puts("It is a trap.");
    57         else printf("%lld
    ", dist[X][1]);
    58     }
    59     return 0; 
    60 }
    View Code

    D. Pants On Fire

    Solved.

    题意:

    给出一些关系,再给出一些关系的询问,对于询问给出三种结果。

    思路:

    Floyd求闭包,但是我还是想知道为什么拓扑排序不行,难道是题意读错,它不一定是个DAG?

     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 
     4 #define N 410
     5 int n, m; 
     6 map <string, int> mp; int cnt; 
     7 char a[N], b[N];
     8 int G[N][N]; 
     9 
    10 int getid(char *s)
    11 {
    12     if (mp.find(s) == mp.end()) mp[s] = ++cnt;
    13     return mp[s];
    14 }
    15 
    16 void Floyd()
    17 {
    18     for (int k = 1; k <= cnt; ++k)
    19     {
    20         for (int i = 1; i <= cnt; ++i)
    21         {
    22             for (int j = 1; j <= cnt; ++j)
    23             {
    24                 if (G[i][k] & G[k][j])
    25                     G[i][j] = 1;
    26             }
    27         }
    28     }
    29 }
    30 
    31 void init()
    32 {
    33     memset(G, 0, sizeof G); 
    34     mp.clear(); cnt = 0; 
    35 }
    36 
    37 int main()
    38 {
    39     while (scanf("%d%d", &n, &m) != EOF)
    40     {
    41         init();
    42         for (int i = 1, u, v; i <= n; ++i)
    43         {
    44             scanf("%s are worse than %s", a, b);
    45             u = getid(a), v = getid(b);
    46             G[u][v] = 1;
    47         }
    48         Floyd(); 
    49         //for (auto it : mp) cout << it.first << " " << rt[it.second] << " " << deep[it.second] << endl;
    50         for (int i = 1, u, v; i <= m; ++i)
    51         {
    52             scanf("%s are worse than %s", a, b);
    53             u = getid(a), v = getid(b);
    54             if (G[u][v] == 1) puts("Fact");
    55             else if (G[v][u] == 1) puts("Alternative Fact");
    56             else puts("Pants on Fire");
    57         }
    58     }
    59     return 0;
    60 }
    View Code

    E. Perpetuum Mobile

    Upsolved.

    题意:

    有n个转换器,m个转换关系,转换关系是有向边

    求从哪个点出发回到起点后经过的转换大于原来的值

    思路:

    枚举起点跑最长路,乘法可以取个$log$, 当然也可以再取个负跑最短路也可以

     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 
     4 #define N 10010
     5 #define INF 0x3f3f3f3f
     6 #define pid pair <int, double>
     7 int n, m;
     8 vector <pid> G[N];
     9 
    10 struct node
    11 {
    12     int to; double w;
    13     node () {}
    14     node (int to, double w) : to(to), w(w) {}
    15     bool operator < (const node &r) const { return w < r.w; }
    16 };
    17 
    18 double dist[N]; bool used[N];
    19 bool Dijkstra(int st)
    20 {
    21     for (int i = 1; i <= n; ++i) dist[i] = 0.0, used[i] = false;
    22     priority_queue <node> q; q.emplace(st, 0);  
    23     while (!q.empty())
    24     {
    25         int u = q.top().to; q.pop(); 
    26         if (used[u] && u == st && dist[u] > 0) return false;  
    27         if (used[u]) continue;
    28         used[u] = 1;
    29         for (auto it : G[u]) if (dist[it.first] < dist[u] + log(it.second)) 
    30         {
    31             dist[it.first] = dist[u] + log(it.second);
    32             q.emplace(it.first, dist[it.first]);
    33         }
    34     }    
    35     return true;
    36 }
    37 
    38 bool solve()
    39 {
    40     for (int i = 1; i <= n; ++i) if (Dijkstra(i) == false)
    41         return false; 
    42     return true;
    43 }
    44 
    45 int main()
    46 {
    47     while (scanf("%d%d", &n, &m) != EOF)
    48     {
    49         for (int i = 1; i <= n; ++i) G[i].clear();
    50         double w;
    51         for (int i = 1, u, v; i <= m; ++i)
    52         {
    53             scanf("%d%d%lf", &u, &v, &w);
    54             G[u].emplace_back(v, w);
    55         }
    56         if (solve() == false) printf("in");
    57         puts("admissible");
    58     }
    59     return 0;
    60 }
    View Code

    F. Plug It In

    Unsolved.

    题意:

    给出一个匹配关系,可以设置一个人和三个人匹配,求最大匹配数。

    思路:

    先求最大匹配,再枚举每个插座,找增广路,枚举点的时候都要复制原图备份一下

     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 
     4 #define N 1510
     5 int n, m, k;
     6 vector <int> G[N];
     7 int linker[N], vis[N], tmp[N];
     8 
     9 bool DFS(int u)
    10 {
    11     for (auto v : G[u]) if (!vis[v])
    12     {
    13         vis[v] = 1; 
    14         if (linker[v] == -1 || DFS(linker[v]))
    15         {
    16             linker[v] = u;
    17             return true;
    18         }
    19     }
    20     return false;
    21 }
    22 
    23 int main()
    24 {
    25     while (scanf("%d%d%d", &n, &m, &k) != EOF)
    26     {
    27         for (int i = 1; i <= n; ++i) G[i].clear();
    28         for (int i = 1, u, v; i <= k; ++i)
    29         {
    30             scanf("%d%d", &u, &v);
    31             G[u].push_back(v);
    32         }
    33         memset(linker, -1, sizeof linker);
    34         int res = 0, ans = 0;
    35         for (int i = 1; i <= n; ++i)
    36         {
    37             memset(vis, 0, sizeof vis);
    38             if (DFS(i)) ++res;
    39         }
    40         memcpy(tmp, linker, sizeof linker);
    41         for (int i = 1; i <= n; ++i)
    42         {
    43             int cnt = 0;
    44             for (int j = 1; j <= 2; ++j)
    45             {
    46                 memset(vis, 0, sizeof vis);
    47                 if (DFS(i)) ++cnt;    
    48             }
    49             ans = max(ans, cnt);
    50             memcpy(linker, tmp, sizeof tmp);
    51         }
    52         printf("%d
    ", res + ans);
    53     }
    54     return 0;
    55 }
    View Code

    G. Water Testing

    Upsolved.

    题意:求一个多边形内部整点个数。

    思路:

    皮克定理:$S = a + frac{b}{2} - 1   S表示多边形面积,frac{b}{2} 表示 多边形边上的整点数   a 表示多边形内部整点数$

    因为点是按顺序给出,有求面积公式$S = frac{1}{2} sum_{i = 0}^{i = n}|vec{P_i} x vec{P_i - 1}|$

    再考虑 线段上的整点个数为

    $令 a = abs(stx - edx), b = abs(sty - edy),整点个数为gcd(a, b) + 1$

    考虑证明:

    根据直线两点式有:

    $y - y_1 = frac{y_2 - y_1}{x_2 - x_1} cdot (x - x _1)$

    移项得

    $y = frac{y_2 - y_1}{x_2 - x_1} cdot (x - x_1) + y_1$

    即考虑 有多少个$x使得 (x - x_1) equiv 0 pmod {x_2 - x_1}$

    由于 $x 在 [x1, x2] 范围内 那么 (x - x_1) 的范围即 [0, x2 - x1]$

    $考虑 frac{y_2 - y_1}{x_2 - x_1} 可以跟他们的gcd约分$

    $frac{x_2 - x_1}{ frac {x_2 - x_1}{gcd((y_2 - y_1), (x_2 - x_1))}} = gcd((y_2 - y_1), (x_2 - x_1))$

     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 
     4 #define N 100010
     5 #define ll long long
     6 int n; 
     7 ll x[N], y[N];
     8 
     9 ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a;}
    10 
    11 int main()
    12 {
    13     while (scanf("%d", &n) != EOF)
    14     {
    15         for (int i = 1; i <= n; ++i)
    16             scanf("%lld%lld", x + i, y + i);
    17         x[0] = x[n]; y[0] = y[n];
    18         ll s = 0, b = 0;
    19         for (int i = 1; i <= n; ++i)
    20         {
    21             s += x[i - 1] * y[i] - x[i] * y[i - 1];
    22             ll A = abs(x[i] - x[i - 1]);
    23             ll B = abs(y[i] - y[i - 1]);
    24             b += gcd(A, B);
    25         }
    26         s = abs(s); 
    27         printf("%lld
    ", (s - b + 2) >> 1);  
    28     }
    29     return 0;
    30 }
    View Code

    H. Ratatoskr

    Unsolved.

    I. Uberwatch

    Water.

     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 
     4 #define N 300010
     5 int n, m, x[N], Max[15];
     6 
     7 int main()
     8 {
     9     while (scanf("%d%d", &n, &m) != EOF)
    10     {
    11         memset(Max, 0, sizeof Max);
    12         for (int i = 1; i <= n; ++i) scanf("%d", x + i);
    13         for (int i = m + 1, res; i <= n; ++i)
    14         {
    15             res = x[i] + Max[m];
    16             for (int j = m; j >= 1; --j) Max[j] = max(Max[j], Max[j - 1]);
    17             Max[1] = max(Max[1], res);
    18         }
    19         printf("%d
    ", *max_element(Max + 1, Max + 1 + m));
    20     }
    21     return 0;
    22 }
    View Code

    J. Word Clock

    Unsolved.

    K. You Are Fired

    Water.

     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 
     4 #define N 10010
     5 int n, k, d;
     6 struct node
     7 {
     8     char s[10]; int c;
     9     void scan()
    10     {
    11         scanf("%s%d", s, &c);
    12     }
    13     bool operator < (const node &r) const {return c > r.c;}
    14 }emp[N];
    15 
    16 int main()
    17 {
    18     while (scanf("%d%d%d", &n, &d, &k) != EOF)
    19     {
    20         for (int i = 1; i <= n; ++i) emp[i].scan();
    21         sort(emp + 1, emp + 1 + n);
    22         int tot = 0;
    23         for (int i = 1; i <= k; ++i) tot += emp[i].c;
    24         if (tot < d) puts("impossible");
    25         else
    26         {
    27             printf("%d
    ", k);
    28             for (int i = 1; i <= k; ++i) printf("%s, YOU ARE FIRED!
    ", emp[i].s);
    29         }
    30     }
    31     return 0;
    32 }
    View Code
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  • 原文地址:https://www.cnblogs.com/Dup4/p/9997377.html
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