hdu 1811 Rank of Tetris

并查集+拓扑排序

把等号的那些东西都用并查集合并一下，这样一来，建立邻接表的时候用根来建立就好了。

然后就是拓扑排序。

如果有两个入度为0的节点，那么说明肯定是条件不足，

如果有成环的肯定是没法排序了。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<algorithm>
using namespace std;

const int maxn = 10010;
const int maxnn = 20010;
int u[maxnn], v[maxnn], s[maxnn][2];
int father[maxn], rudu[maxn], ff[maxn], yy[maxn];
vector<int>ljb[maxn];

int find(int x)
{
    while (father[x] != x) x = father[x] = father[father[x]];
    return father[x];
}

int main()
{
    int n, m, i, ii, j;
    while (~scanf("%d%d", &n, &m))
    {
        int jieguo = 1;
        for (i = 0; i <= n; i++) father[i] = i;
        memset(rudu, 0, sizeof(rudu));
        memset(ff, 0, sizeof(ff));
        for (i = 0; i <= n; i++) ljb[i].clear();
        for (i = 0; i < m; i++) scanf("%d%s%d", &u[i], s[i], &v[i]);
        for (i = 0; i < m; i++)
        {
            if (s[i][0] == '=')
            {
                int fu = find(u[i]);
                int fv = find(v[i]);
                father[fu] = fv;
            }
        }
        int tott = 0;
        for (i = 0; i < n; i++)
        {
            int gen = find(i);
            if (ff[gen] == 0){ yy[tott] = gen; tott++; ff[gen] = 1; }
        }
        for (i = 0; i < m; i++)
        {
            if (s[i][0] == '>')
            {
                int fu = find(u[i]);
                int fv = find(v[i]);
                ljb[fu].push_back(fv);
                rudu[fv]++;
            }
            else if (s[i][0] == '<')
            {
                int fu = find(u[i]);
                int fv = find(v[i]);
                ljb[fv].push_back(fu);
                rudu[fu]++;
            }
        }
        int summ, r, cun[maxn], tongji = 0;
        while (1)
        {
            if (tongji == tott) break;
            summ = 0;
            for (i = 0; i < tott; i++)
                if (rudu[yy[i]] == 0)
                    cun[summ] = yy[i], summ++, r = yy[i];    
            if (summ !=0) 
            { 
                tongji = tongji + summ;
                if (summ >= 2)jieguo = 3; 
                for (i = 0; i < summ; i++)
                {
                    rudu[cun[i]]--;
                    for (j = 0; j < ljb[cun[i]].size(); j++)
                        rudu[ljb[cun[i]][j]]--;
                }
            }
            if (summ == 0){ jieguo = 2; break; }    
        }
        if (jieguo == 1) printf("OK\n");
        else if (jieguo == 2) printf("CONFLICT\n");
        else if (jieguo == 3) printf("UNCERTAIN\n");
    }
    return 0;
}