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

*/