c/c++语言开发共享BZOJ3498: PA2009 Cakes(三元环)

题意

题目链接

sol

按照套路把边转成无向图，我们采取的策略是从权值大的向权值小的连边

然后从按权值从小到大枚举每个点，再枚举他们连出去的点(v)

如果(v)的度数(leqslant m)，那么就再暴力枚举(v)连出去的点(t)，看(u)与(t)是否联通(打标记)

否则暴力枚举(u)连出去的点(t)，看(v)与(t)是否联通(直接hash表)

复杂度为(o(m sqrt{m}))

 #include<bits/stdc++.h> #define ll long long  using namespace std; const int maxn = 100001; inline int read() {     char c = getchar(); int x = 0, f = 1;     while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}     while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();     return x * f; } int n, m, a[maxn], block, siz[maxn], flag[maxn], rak[maxn], tp[maxn]; vector<int> v[maxn]; set<int> s[maxn]; int comp(const int &x, const int &y) {     return a[x] == a[y] ? x < y : a[x] < a[y]; } int main() {     n = read(); m = read(); block = sqrt(m);     for(int i = 1; i <= n; i++) a[i] = read(), tp[i] = i;     sort(tp + 1, tp + n + 1, comp);     for(int i = 1; i <= n; i++) rak[tp[i]] = i;      for(int i = 1; i <= m; i++) {         int x = read(), y = read();         if(rak[x] > rak[y]) v[x].push_back(y), siz[x]++;         else v[y].push_back(x), siz[y]++;     }     ll ans = 0;     for(int i = 3; i <= n; i++) {         int x = tp[i];         for(int j = 0, to; j < v[x].size(); j++) flag[to = v[x][j]] = i;         for(int j = 0, to; j < v[x].size(); j++) {             if(siz[to = v[x][j]] <= block) {                 for(int k = 0; k < v[to].size(); k++)                      if(flag[v[to][k]] == i) ans += a[x];             } else {                 for(int k = 0; k < v[x].size(); k++)                     if(s[to].count(v[x][k])) ans += a[x];             }             s[x].insert(to);         }     }     cout << ans;     return 0; } /* 2 1 13 17 2 1  */