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  • BZOJ 3498: PA2009 Cakes 一类经典的三元环计数问题

    首先引入一个最常见的经典三元环问题。

    #include <bits/stdc++.h>
    using namespace std;
    const int maxn = 100005;
    vector <int> g[maxn], low, high;
    map <int, int> mp[maxn];
    int n, m, in[maxn], vis[maxn];
    
    int main()
    {
        scanf("%d %d", &n,&m);
        for(int i=1; i<=m; i++){
            int x,y;
            scanf("%d %d", &x,&y);
            g[x].push_back(y);
            g[y].push_back(x);
            mp[x][y] = 1;
            mp[y][x] = 1;
            in[x]++;
            in[y]++;
        }
        for(int i=1; i<=n; i++){
            if(in[i]<=sqrt(m)) low.push_back(i);
            else high.push_back(i);
        }
        int ans = 0;
        memset(vis, false, sizeof(vis));
        for(int i=0; i<low.size(); i++){
            int x = low[i];
            vis[x] = 1;
            for(int j=0; j<g[x].size(); j++){
                int y = g[x][j];
                if(vis[y]) continue;
                for(int k=j+1; k<g[x].size(); k++){
                    int z = g[x][k];
                    if(vis[z]) continue;
                    if(mp[y].count(z)) ans++;
                }
            }
        }
        for(int i=0; i<high.size(); i++){
            for(int j=i+1; j<high.size(); j++){
                int x, y, z;
                x = high[i];
                y = high[j];
                if(mp[x].count(y)==0) continue;
                for(int k=j+1; k<high.size(); k++){
                    z = high[k];
                    if(mp[y].count(z)&&mp[x].count(z)) ans++;
                }
            }
        }
        printf("%d
    ", ans);
        return 0;
    }
    

     BZOJ 3948

    //BZOJ 3498
    #include <bits/stdc++.h>
    using namespace std;
    const int maxn = 100005;
    typedef long long LL;
    struct node{
        int val,id;
        node(){}
        node(int val,int id):val(val),id(id){}
        bool operator<(const node &rhs)const{
            return val>rhs.val;
        }
    }a[maxn];
    vector <int> G[maxn];
    map <int, int> mp[maxn];
    int Rank[maxn], out[maxn], linker[maxn];
    int n, m;
    
    int main()
    {
        scanf("%d %d", &n,&m);
        for(int i=1; i<=n; i++){
            scanf("%d", &a[i].val);
            a[i].id = i;
        }
        sort(a+1, a+n+1);
        for(int i=1; i<=n; i++) Rank[a[i].id] = i;
        for(int i=1; i<=m; i++){
            int x, y;
            scanf("%d %d", &x,&y);
            if(Rank[x]<Rank[y]){
                G[x].push_back(y), out[x]++, mp[x][y]=1;
            }else{
                G[y].push_back(x), out[y]++, mp[y][x]=1;
            }
        }
        LL ans = 0;
        for(int i=1; i<=n; i++){
            int x = a[i].id, y;
            for(int j=0; j<G[x].size(); j++){
                linker[G[x][j]] = x;
            }
            for(int j=0; j<G[x].size(); j++){
                y = G[x][j];
                if(out[y]>sqrt(m)+1){
                    for(int k=0; k<G[x].size(); k++){
                        int z = G[x][k];
                        if(mp[y].count(z)) ans += a[i].val;
                    }
                }else{
                    for(int k=0; k<G[y].size(); k++){
                        int z = G[y][k];
                        if(linker[z] == x) ans += a[i].val;
                    }
                }
            }
        }
        printf("%lld
    ", ans);
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/spfa/p/7495438.html
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