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  • 比赛-Round (10 Jul)

    1. 无线通讯网

     1 #include <stdio.h>
     2 #include <algorithm>
     3 #include <math.h>
     4 
     5 using namespace std;
     6 
     7 typedef long long LL;
     8 
     9 const int _N = 600;
    10 
    11 LL P, S;
    12 LL Dad[_N], dis[_N][_N], A[_N*_N], X[_N], Y[_N];
    13 bool mk[_N];
    14 
    15 LL _fun(LL val) { return val * val; }
    16 
    17 LL GetDad(LL v) { return Dad[v] == v ? v : Dad[v] = GetDad(Dad[v]); }
    18 
    19 bool Check(LL lim)
    20 {
    21     LL i, j;
    22     for (i = 1; i <= P; ++i) Dad[i] = i, mk[i] = false;
    23     for (i = 1; i <= P; ++i)
    24         for (j = 1; j <= P; ++j)
    25             if (i != j && dis[i][j] <= A[lim]) {
    26                 LL t1 = GetDad(i), t2 = GetDad(j);
    27                 if (t1 != t2) Dad[t1] = t2;
    28             }
    29     LL cnt = 0;
    30     for (i = 1; i <= P; ++i) {
    31         LL tmp = GetDad(i);
    32         if (!mk[tmp]) mk[tmp] = true, ++cnt;
    33     }
    34     return cnt <= S;
    35 }
    36 
    37 int main()
    38 {
    39     freopen("wireless.in", "r", stdin);
    40     freopen("wireless.out", "w", stdout);
    41     LL i, j, Acnt = 0;
    42     scanf("%lld%lld", &S, &P);
    43     for (i = 1; i <= P; ++i) scanf("%lld%lld", &X[i], &Y[i]);
    44     for (i = 1; i <= P; ++i)
    45         for (j = 1; j <= P; ++j)
    46             if (i != j) A[++Acnt] = dis[i][j] = _fun(X[i]-X[j]) + _fun(Y[i]-Y[j]);
    47     A[++Acnt] = 0;
    48     sort(A+1, A+1+Acnt);
    49     LL l = 1, r = Acnt;
    50     while (l <= r) {
    51         LL mid = l+r >> 1;
    52         if (Check(mid)) r = mid - 1;
    53         else l = mid + 1;
    54     }
    55     printf("%.2lf
    ", sqrt(A[l]));
    56     
    57     return 0;
    58 }
    View Code

    2. 混合图

     1 #include <stdio.h>
     2 #include <vector>
     3 #include <queue>
     4 
     5 using namespace std;
     6 
     7 const int _N = 120000;
     8 
     9 queue<int> Q;
    10 vector<int> G[_N];
    11 
    12 int ind[_N], pos[_N];
    13 
    14 inline void Ins(int a, int b) { G[a].push_back(b); ++ind[b]; return; }
    15 
    16 int main()
    17 {
    18     int N, i, M1, M2, a, b;
    19     scanf("%d%d%d", &N, &M1, &M2);
    20     for (i = 1; i <= M1; ++i)
    21         scanf("%d%d", &a, &b), Ins(a, b);
    22     for (i = 1; i <= N; ++i) {
    23         if (!ind[i]) Q.push(i);
    24         pos[i] = N+1;
    25     }
    26     int cnt = 0;
    27     while (!Q.empty()) {
    28         int t = Q.front();
    29         pos[t] = ++cnt;
    30         Q.pop();
    31         for (i = G[t].size()-1; i >= 0; --i)
    32             if(!--ind[G[t][i]]) Q.push(G[t][i]);
    33     }
    34     for (i = 1; i <= M2; ++i) {
    35         scanf("%d%d", &a, &b);
    36         if (pos[a] > pos[b]) swap(a, b);
    37         printf("%d %d
    ", a, b);
    38     }
    39     return 0;
    40 }
    View Code

    3. 小K的农场

    非常重要。学到了新操作:SPFA 的 SLF, LLL 优化,以及在判负环时的 DFS 优化,还有权值无意义时(只判负环时)的 DFS, dis[i] = 0 优化。

    详细见 SPFA 模板那篇随笔(http://www.cnblogs.com/ghcred/p/8051261.html)

     1 #include <stdio.h>
     2 #include <queue>
     3 
     4 using namespace std;
     5 
     6 const int _N = 12000;
     7 const int INF = 1e9;
     8 
     9 struct edge {
    10     int v, w;
    11     edge(int v, int w):
    12         v(v), w(w) { }
    13 };
    14 
    15 vector<edge> G[_N];
    16 int N, M;
    17 int dis[_N], cnt[_N];
    18 bool mk[_N];
    19 deque<int> Q;
    20 
    21 bool SPFA(int p)
    22 {
    23     bool flag = false;
    24     mk[p] = true;
    25     for (int i = G[p].size()-1; i >= 0; --i) {
    26         edge v = G[p][i];
    27         if (dis[v.v] > dis[p]+v.w) {
    28             dis[v.v] = dis[p]+v.w;
    29             if (!mk[v.v]) {
    30                 if (SPFA(v.v)) { flag = true; break; }
    31             } else {
    32                 flag = true;
    33                 break;
    34             }
    35         }
    36     }
    37     mk[p] = false;
    38     return flag;
    39 }
    40 
    41 void Ins(int t0, int t1, int t2) { G[t0].push_back(edge(t1, t2)); return; }
    42 
    43 int main()
    44 {
    45     int i;
    46     scanf("%d%d", &N, &M);
    47     for (i = 1; i <= M; ++i) {
    48         int a, b, c, ins;
    49         scanf("%d%d%d", &ins, &a, &b);
    50         if (ins == 1)
    51             scanf("%d", &c), Ins(b, a, c);
    52         else if (ins == 2)
    53             scanf("%d", &c), Ins(a, b, -c);
    54         else
    55             Ins(a, b, 0), Ins(b, a, 0);
    56     }
    57     for (i = 1; i <= N; ++i)
    58         if (SPFA(i)) break;
    59     if (i != N+1) printf("No
    ");
    60     else printf("Yes
    ");
    61     return 0;
    62 }
    View Code
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  • 原文地址:https://www.cnblogs.com/ghcred/p/9288365.html
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