hdu5486 Difference of Clustering 暴力

这题什么情况。。。。。直接暴力加标记,复杂度o(n)啊。。。明显能过啊。。。明显是除了签到最简单的一道。。。。居然过这么少人。。。估计很多人看到merge和split以为是什么数据结构。。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<vector>
#include<map>
#define REP(i,a,b) for(int i=a;i<=b;i++)
#define MS0(a) memset(a,0,sizeof(a))
#define key_val ch[ch[rt[i]][1]][0]
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

using namespace std;

typedef long long ll;
const int maxn=1000100;
const int INF=1e9+10;

struct Edge
{
    int u,v;
    friend bool operator<(Edge A,Edge B)
    {
        return A.u==B.u?A.v<B.v:A.u<B.u;
    }
};Edge e[maxn];int m;
vector<int> G[maxn];
int un,vn;
map<int,int> id,id2;
bool vis[maxn];
int deg[maxn];

void dfs(int u)
{
    if(vis[u]) return;
    vis[u]=1;
    for(int i=0;i<G[u].size();i++){
        dfs(G[u][i]);
    }
}

int main()
{
    #ifndef ONLINE_JUDGE
        freopen("in.txt","r",stdin);
    #endif
    int T;cin>>T;
    REP(casen,1,T){
        scanf("%d",&m);
        REP(i,1,m) scanf("%d%d",&e[i].u,&e[i].v),e[i].u++,e[i].v++;
        sort(e+1,e+m+1);int tm=m;m=1;
        REP(i,2,tm) if(e[i].u!=e[i-1].u||e[i].v!=e[i-1].v) e[++m]=e[i];
        id.clear();un=0;
        REP(i,1,m){
            if(id[e[i].u]) continue;
            id[e[i].u]=++un;
        }
        vn=un;
        id2.clear();vn=un;
        REP(i,1,m){
            if(id2[e[i].v]) continue;
            id2[e[i].v]=++vn;
        }
        REP(i,1,vn) G[i].clear();MS0(deg);
        REP(i,1,m){
            e[i].u=id[e[i].u];
            e[i].v=id2[e[i].v];
            G[e[i].u].push_back(e[i].v);
            G[e[i].v].push_back(e[i].u);
            deg[e[i].u]++;
            deg[e[i].v]++;
        }
        int A=0,B=0,C=0;/// 1-n n-1 1-1
        MS0(vis);
        int v;
        //cout<<"un="<<un<<" vn="<<vn<<endl;
        //REP(i,1,vn) cout<<deg[i]<<" ";cout<<endl;
        REP(i,1,un){
            if(vis[i]) continue;
            if(deg[i]==1){
                v=G[i][0];
                if(deg[v]==1) C++;
                else dfs(i);
            }
            else if(deg[i]>1){
                int tag=1;
                for(int j=0;j<G[i].size();j++){
                    v=G[i][j];
                    if(deg[v]!=1){
                        tag=0;break;
                    }
                }
                if(tag==0) dfs(i);
                else A++;
            }
        }
        MS0(vis);
        REP(i,un+1,vn){
            if(vis[i]) continue;
            if(deg[i]==1){
                v=G[i][0];
                if(deg[v]!=1) dfs(i);
            }
            else if(deg[i]>1){
                int tag=1;
                for(int j=0;j<G[i].size();j++){
                    v=G[i][j];
                    if(deg[v]!=1){
                        tag=0;break;
                    }
                }
                if(tag==0) dfs(i);
                else B++;
            }
        }
        printf("Case #%d: %d %d %d
",casen,A,B,C);
    }
    return 0;
}
View Code
没有AC不了的题,只有不努力的ACMER!
原文地址:https://www.cnblogs.com/--560/p/5211638.html