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hdu 1811 Rank of Tetris

2015-06-04 14:28 363 查看
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;
}
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