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\n", 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\n", ans);
    return 0;
}

 

posted @ 2017-09-08 16:55  zxycoder  阅读(898)  评论(0编辑  收藏  举报