2017中国大学生程序设计竞赛 - 网络选拔赛 HDU 6152 Friend-Graph 暴暴暴暴力

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6152

题意:判定一个无向图是否有三个点的团或者三个点的独立集。

解法:Ramsey theorem,n >= 6 直接输出 Bad 否则暴力。我是直接暴力,加个break优化就好了。

 

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 3e3+3;
bool w[maxn][maxn];
bool check(int n){
    for(int i=1; i<=n; i++){
        for(int j=i+1; j<=n; j++){
            for(int k=j+1; k<=n; k++){
                if(w[i][j]==w[i][k]&&w[i][k]==w[j][k]&&w[i][j]==w[j][k]){
                    return true;
                }
            }
        }
    }
    return false;
}
struct FastIO
{
    static const int S = 1310720;
    int wpos;
    char wbuf[S];
    FastIO() : wpos(0) {}
    inline int xchar()
    {
        static char buf[S];
        static int len = 0, pos = 0;
        if(pos == len)
            pos = 0, len = fread(buf, 1, S, stdin);
        if(pos == len)
            exit(0);
        return buf[pos ++];
    }
    inline unsigned long long xuint()
    {
        int c = xchar();
        unsigned long long x = 0;
        while(c <= 32)
            c = xchar();
        for(; '0' <= c && c <= '9'; c = xchar())
            x = x * 10 + c - '0';
        return x;
    }
    inline long long xint()
    {
        long long s = 1;
        int c = xchar(), x = 0;
        while(c <= 32)
            c = xchar();
        if(c == '-')
            s = -1, c = xchar();
        for(; '0' <= c && c <= '9'; c = xchar())
            x = x * 10 + c - '0';
        return x * s;
    }
    inline void xstring(char *s)
    {
        int c = xchar();
        while(c <= 32)
            c = xchar();
        for(; c > 32; c = xchar())
            * s++ = c;
        *s = 0;
    }
    inline double xdouble()
    {
        bool sign = 0;
        char ch = xchar();
        double x = 0;
        while(ch <= 32)
            ch = xchar();
        if(ch == '-')
            sign = 1, ch = xchar();
        for(; '0' <= ch && ch <= '9'; ch = xchar())
            x = x * 10 + ch - '0';
        if(ch == '.')
        {
            double tmp = 1;
            ch = xchar();
            for(; ch >= '0' && ch <= '9'; ch = xchar())
                tmp /= 10.0, x += tmp * (ch - '0');
        }
        if(sign)
            x = -x;
        return x;
    }
    inline void wchar(int x)
    {
        if(wpos == S)
            fwrite(wbuf, 1, S, stdout), wpos = 0;
        wbuf[wpos ++] = x;
    }
    inline void wint(long long x)
    {
        if(x < 0)
            wchar('-'), x = -x;
        char s[24];
        int n = 0;
        while(x || !n)
            s[n ++] = '0' + x % 10, x /= 10;
        while(n--)
            wchar(s[n]);
    }
    inline void wstring(const char *s)
    {
        while(*s)
            wchar(*s++);
    }
    inline void wdouble(double x, int y = 6)
    {
        static long long mul[] = {1, 10, 100, 1000, 10000, 100000, 1000000, 10000000, 100000000, 1000000000, 10000000000LL, 100000000000LL, 1000000000000LL, 10000000000000LL, 100000000000000LL, 1000000000000000LL, 10000000000000000LL, 100000000000000000LL};
        if(x < -1e-12)
            wchar('-'), x = -x;
        x *= mul[y];
        long long x1 = (long long) floorl(x);
        if(x - floor(x) >= 0.5)
            ++x1;
        long long x2 = x1 / mul[y], x3 = x1 - x2 * mul[y];
        wint(x2);
        if(y > 0)
        {
            wchar('.');
            for(size_t i = 1; i < y && x3 * mul[i] < mul[y]; wchar('0'), ++i);
            wint(x3);
        }
    }
    ~FastIO()
    {
        if(wpos)
            fwrite(wbuf, 1, wpos, stdout), wpos = 0;
    }
} io;

int main()
{
    int T,n,m;
    T = io.xint();
    while(T--)
    {
        n = io.xint();
        for(int i=1; i<n; i++){
            for(int j=i+1; j<=n; j++){
                m = io.xint();
                if(m == 1) w[i][j]=w[j][i]=false;
                else w[i][j]=w[j][i]=true;
            }
        }
        if(check(n)){
            puts("Bad Team!");
        }
        else{
            puts("Great Team!");
        }
    }
    return 0;
}

 

posted @ 2017-08-19 19:57  zxycoder  阅读(193)  评论(0编辑  收藏  举报