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;

}