題意

題目鏈接

Sol

按照套路把邊轉成無向圖，我們采取的策略是從權值大的向權值小的連邊

然後從按權值從小到大枚舉每個點，再枚舉他們連出去的點\(v\)

如果\(v\)的度數\(\leqslant M\)，那麽就再暴力枚舉\(v\)連出去的點\(t\)，看\(u\)與\(t\)是否聯通(打標記)

否則暴力枚舉\(u\)連出去的點\(t\)，看\(v\)與\(t\)是否聯通(直接hash表)

復雜度為\(O(M \sqrt{M})\)

#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

*/
 

BZOJ3498: PA2009 Cakes(三元環)