• POJ 1703 Find them, Catch them


    简单带权并查集0,1关系

    //#pragma comment(linker, "/STACK:1024000000,1024000000")
    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<algorithm>
    #include<iostream>
    #include<sstream>
    #include<cmath>
    #include<climits>
    #include<string>
    #include<map>
    #include<queue>
    #include<vector>
    #include<stack>
    #include<set>
    using namespace std;
    typedef long long ll;
    typedef pair<int,int> pii;
    #define pb(a) push_back(a)
    #define INF 0x1f1f1f1f
    #define lson idx<<1,l,mid
    #define rson idx<<1|1,mid+1,r
    #define PI  3.1415926535898
    template<class T> T min(const T& a,const T& b,const T& c) {
        return min(min(a,b),min(a,c));
    }
    template<class T> T max(const T& a,const T& b,const T& c) {
        return max(max(a,b),max(a,c));
    }
    void debug() {
    #ifdef ONLINE_JUDGE
    #else
    
        freopen("d:\in.txt","r",stdin);
        freopen("d:\out1.txt","w",stdout);
    #endif
    }
    int getch() {
        int ch;
        while((ch=getchar())!=EOF) {
            if(ch!=' '&&ch!=' ')return ch;
        }
        return EOF;
    }
    int da[110000],rel[110000];
    int find(int a)
    {
        if(a==da[a])return a;
        int root=find(da[a]);
        rel[a]=(rel[a]+rel[da[a]])%2;
        return da[a]=root;
    }
    int main()
    {
        int t;
        scanf("%d",&t);
        for(int ca=1;ca<=t;ca++)
        {
            int n,q;
            scanf("%d%d",&n,&q);
            for(int i=1;i<=n;i++)
            {
                da[i]=i;
                rel[i]=0;
            }
            for(int Q=1;Q<=q;Q++)
            {
                char op[10];
                int a,b;
                scanf("%s%d%d",op,&a,&b);
                int fa=find(a),fb=find(b);
                if(op[0]=='A')
                {
                    if(fa!=fb)
                        printf("Not sure yet.
    ");
                    else printf("%s
    ",(rel[a]-rel[b]+2)%2?"In different gangs.":"In the same gang.");
                }else
                {
                    if(fa!=fb)
                    {
                        da[fb]=fa;
                        rel[fb]=(rel[a]-rel[b]+1)%2;
                    }
                }
            }
        }
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/BMan/p/3274849.html
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